CHAPTER 4 CHEMICAL REACTIONS PRACTICE EXAMPLES 1A
(E) (a)
(b)
1B
(E) (a)
(b)
2A
Unbalanced reaction: Balance Ca & PO43-: Balance H atoms: Self Check:
H3PO4(aq) + CaO(s) 2 H3PO4(aq) + 3 CaO(s) 2 H3PO4(aq) + 3 CaO(s) 6 H + 2 P + 11 O + 3 Ca
Ca3(PO4)2(aq) + H2O(l) Ca3(PO4)2(aq) + H2O(l) Ca3(PO4)2(aq) + 3 H2O(l) 6 H + 2 P + 11 O + 3 Ca
Unbalanced reaction: Balance C& H: Balance O atoms: Self Check:
C3H8(g) + O2(g) C3H8(g) + O2(g) C3H8(g) + 5 O2(g) 3 C + 8 H + 10 O
CO2(g) + H2O(g) 3 CO2(g) + 4 H2O(g) 3 CO2(g) + 4 H2O(g) 3 C + 8 H + 10 O
Unbalanced reaction: NH3(g) + O2(g) Balance N and H: NH3(g) + O2(g) Balance O atoms: NH3(g) + 7/4 O2(g) Multiply by 4 (whole #): 4 NH3(g) + 7 O2(g) Self Check: 4 N + 12 H + 14 O
NO2(g) + H2O(g) NO2(g) + 3/2 H2O(g) NO2(g) + 3/2 H2O(g) 4 NO2(g) + 6 H2O(g) 4 N + 12 H + 14 O
Unbalanced reaction: Balance H atoms: Balance O atoms: Balance N atoms: Multiply by 4 (whole #) Self Check:
NO2(g) + NH3(g) NO2(g) + 2 NH3(g) 3/2 NO2(g) + 2 NH3(g) 3/2 NO2(g) + 2 NH3(g) 6 NO2(g) + 8 NH3(g) 14 N + 24 H + 12 O
N2(g) + H2O(g) N2(g) + 3 H2O(g) N2(g) + 3 H2O(g) 7/4 N2(g) + 3 H2O(g) 7 N2(g) + 12 H2O(g) 14 N + 24 H + 12 O
HgS(s) + CaO(s) HgS(s) + 4 CaO(s) HgS(s) + 4 CaO(s) 4 HgS(s) + 4 CaO(s) 4 HgS(s) + 4 CaO(s) 4 Hg + 4 S + 4 O + 4 Ca
(E) Unbalanced reaction: Balance O atoms: Balance Ca atoms: Balance S atoms: Balance Hg atoms: Self Check:
112
CaS(s) + CaSO4(s) + Hg(l) CaS(s) + CaSO4(s) + Hg(l) 3 CaS(s) + CaSO4(s) + Hg(l) 3 CaS(s) + CaSO4(s) + Hg(l) 3 CaS(s) + CaSO4(s) + 4 Hg(l) 4 Hg + 4 S + 4 O + 4 Ca
Chapter 4: Chemical Reactions
2B
(E) Unbalanced reaction: Balance C atoms: Balance S atoms: Balance H atoms: Balance O atoms: Multiply by 2 (whole #): Self Check:
3A
(E) The balanced chemical equation provides the factor needed to convert from moles 3 mol O 2 KClO 3 to moles O2. Amount O 2 = 1.76 mol KClO 3 = 2.64 mol O 2 2 mol KClO 3
3B
(E) First, find the molar mass of Ag 2 O . 2 mol Ag 107.87 g Ag +16.00 g O = 231.74 g Ag 2 O / mol
CO2(g) + H2O(l) + SO2(g) C7H6O2S(l) + O2 (g) C7H6O2S(l) + O2(g) 7 CO2(g) + H2O(l) + SO2(g) C7H6O2S(l) + O2(g) 7 CO2(g) + H2O(l) + SO2(g) 7 CO2(g) + 3 H2O(l) + SO2(g) C7H6O2S(l) + O2(g) C7H6O2S(l) + 8.5 O2 (g) 7 CO2(g) + 3 H2O(l) + SO2(g) 2 C7H6O2S(l) + 17 O2 (g) 14 CO2(g) + 6 H2O(l) + 2 SO2(g) 14 C + 12 H + 2 S + 38 O 14 C + 12 H + 2 S + 38 O
amount Ag = 1.00 kg Ag 2 O 4A
(E) The balanced chemical equation provides the factor to convert from amount of Mg to amount of Mg 3 N 2 . First we must determine the molar mass of Mg 3 N 2 . molar mass = 3mol Mg 24.305g Mg + 2 mol N 14.007 g N = 100.93g Mg 3 N 2
mass Mg 3 N 2 = 3.82g Mg 4B
1000 g 1 mol Ag 2 O 2 mol Ag = 8.63 mol Ag 1.00 kg 231.74 g Ag 2 O 1 mol Ag 2 O
1mol Mg 1mol Mg 3 N 2 100.93g Mg 3 N 2 = 5.29 g Mg 3 N 2 24.31g Mg 3mol Mg 1mol Mg 3 N 2
(E) The pivotal conversion is from H 2 g to CH 3OH (l). For this we use the balanced
equation, which requires that we use the amounts in moles of both substances. The solution involves converting to and from amounts, using molar masses. 2 mol H 2 2.016 g H 2 1000 g 1mol CH 3 OH mass H 2 g = 1.00 kg CH 3 OH(l) 1 kg 32.04 g CH 3 OH 1mol CH 3 OH 1mol H 2 mass H 2 g = 126 g H 2
5A
(M) The equation for the cited reaction is: 2 NH 3 g + 1.5 O 2 g N 2 (g) + 3H 2 O l
The pivotal conversion is from one substance to another, in moles, with the balanced chemical equation providing the conversion factor. 2 mol NH 3 17.0305g NH 3 1mol O 2 mass NH 3 g = 1.00 g O 2 g = 0.710 g NH 3 32.00 g O 2 1.5 mol O 2 1mol H 2
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Chapter 4: Chemical Reactions
5B
(M) The equation for the combustion reaction is: 25 C8 H18 l + O 2 g 8 CO 2 g + 9 H 2 O l 2 1mol C8 H18 12.5 mol O 2 32.00 g O 2 mass O 2 = 1.00 g C8 H18 = 3.50 g O 2 g 114.23g C8 H18 1mol C8 H18 1mol O 2
6A
(M) We must convert mass H 2 amount of H 2 amount of Al mass of Al mass of alloy volume of alloy. The calculation is performed as follows: each arrow in the preceding sentence requires a conversion factor. 1mol H 2 2 mol Al 26.98g Al 100.0 g alloy 1cm3alloy Valloy 1.000 g H 2 2.016 g H 2 3mol H 2 1mol Al 93.7 g Al 2.85g alloy
6B
7A
Volume of alloy = 3.34 cm3 alloy (M) In the example, 0.207 g H 2 is collected from 1.97 g alloy; the alloy is 6.3% Cu by mass. This information provides the conversion factors we need. 1.97 g alloy 6.3 g Cu mass Cu = 1.31g H 2 = 0.79 g Cu 0.207 g H 2 100.0 g alloy Notice that we do not have to consider each step separately. We can simply use values produced in the course of the calculation as conversion factors.
(M) The cited reaction is 2 Al s + 6 HCl aq 2 AlCl3 aq + 3 H 2 g . The HCl(aq)
solution has a density of 1.14 g/mL and contains 28.0% HCl. We need to convert between the substances HCl and H 2 ; the important conversion factor comes from the balanced chemical equation. The sequence of conversions is: volume of HCl(aq) mass of HCl(aq) mass of pure HCl amount of HCl amount of H 2 mass of H 2 . In the calculation below, each arrow in the sequence is replaced by a conversion factor. 1.14 g sol 28.0 g HCl 1mol HCl 3mol H 2 2.016 g H 2 mass H 2 = 0.05 mL HCl aq 1mL soln 100.0 g soln 36.46 g HCl 6 mol HCl 1mol H 2 mass H 2 = 4 104 g H 2 g = 0.4 mg H 2 g 7B
(M) Density is necessary to determine the mass of the vinegar, and then the mass of acetic acid. mass CO 2 (g) = 5.00 mL vinegar×
1mol CO2 44.01g CO 2 1.01g 0.040g acid 1mol CH3COOH × × × × 1mL 1g vinegar 60.05g CH3COOH 1mol CH3COOH 1mol CO2
= 0.15g CO2
8A
(M) Determine the amount in moles of acetone and the volume in liters of the solution. 22.3g CH 3 2 CO× molarity of acetone =
1mol CH 3 2 CO
58.08g CH 3 2 CO
1.25 L soln
114
= 0.307 M
Chapter 4: Chemical Reactions
8B
(M) The molar mass of acetic acid, HC 2 H 3O 2 , is 60.05 g/mol. We begin with the quantity of acetic acid in the numerator and that of the solution in the denominator, and transform to the appropriate units for each. molarity =
15.0 mL HC 2 H 3O 2 1000 mL 1.048g HC 2 H 3O 2 1mol HC 2 H 3O 2 = 0.524 M 500.0 mL soln 1L soln 1mL HC 2 H 3O 2 60.05g HC 2 H 3O 2
9A
(E) The molar mass of NaNO 3 is 84.99 g/mol. We recall that “M” stands for “mol /L soln.” 10.8 mol NaNO3 84.99 g NaNO3 1L mass NaNO3 = 125 mL soln = 115 g NaNO3 1000 mL 1L soln 1mol NaNO3
9B
(E) We begin by determining the molar mass of Na 2SO 4 10H 2 O . The amount of solute needed is computed from the concentration and volume of the solution. mass Na 2SO 4 10H 2 O = 355 mL soln
1L 1000 mL
322.21 g Na 2SO 4 10H 2 O 1 mol Na 2SO 4 10H 2 O
0.445 mol Na 2SO 4 1 L soln
1 mol Na 2SO 4 10H 2 O 1 mol Na 2SO 4
50.9 g Na 2SO 4 10H 2 O
10A (E) The amount of solute in the concentrated solution doesn’t change when the solution is diluted. We take advantage of an alternative definition of molarity to answer the question: millimoles of solute/milliliter of solution. 0.450 mmol K 2 CrO 4 amount K 2 CrO 4 = 15.00 mL = 6.75 mmol K 2 CrO 4 1 mL soln 6.75 mmol K 2 CrO 4 K 2 CrO 4 molarity, dilute solution = = 0.0675 M 100.00 mL soln 10B (E) We know the initial concentration (0.105 M) and volume (275 mL) of the solution, along with its final volume (237 mL). The final concentration equals the initial concentration times a ratio of the two volumes. V 275mL cf ci i 0105 . M 0122 . M Vf 237 mL 11A (M) The balanced equation is K 2 CrO 4 aq 2 AgNO3 aq Ag 2 CrO 4 s 2 KNO3 aq .
The molar mass of Ag 2 CrO 4 is 331.73 g mol . The conversions needed are mass Ag 2 CrO 4 amount Ag 2 CrO 4 (moles) amount K 2 CrO 4 (moles) volume K 2 CrO 4 aq . VK 2CrO4 1.50 g Ag 2 CrO 4
1mol Ag 2 CrO 4
331.73g Ag 2CrO 4 1000 mL solution =18.1mL 1L solution
115
1mol K 2CrO 4 1L soln 1 mol Ag 2 CrO 4 0.250 mol K 2 CrO 4
Chapter 4: Chemical Reactions
11B (M) Balanced reaction: 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq) moles of K2CrO4 = C V = 0.0855 M 0.175 L sol = 0.01496 moles K2CrO4 2 mol AgNO3 = 0.0299 mol AgNO3 moles of AgNO3 = 0.01496 mol K2CrO4 1 mol K 2 CrO 4 0.0299 mol AgNO3 n = 0.1995 L or 2.00 102 mL (0.200 L) of AgNO3 VAgNO3 = = C 0.150 mol AgNO 3 L 1mol Ag 2 CrO 4 331.73 g Ag 2 CrO 4 Mass of Ag2CrO4 formed = 0.01496 moles K2CrO4 1 mol K 2 CrO 4 1mol Ag 2 CrO 4 Mass of Ag2CrO4 formed = 4.96 g Ag2CrO4 12A (M) Reaction: P4 s 6 Cl2 g 4 PCl3 l . We must determine the mass of PCl 3 formed
by each reactant. 1 mol P4 4 mol PCl 3 137.33 g PCl 3 953 g PCl 3 123.90g P4 1mol P4 1 mol PCl 3 1 mol Cl 2 4 mol PCl 3 137.33 g PCl 3 mass PCl 3 725 g Cl 2 936 g PCl 3 70.91g Cl 2 6 mol Cl 2 1 mol PCl 3 Thus, a maximum of 936g PCl 3 can be produced; there is not enough Cl 2 to produce any more. mass PCl 3 215 g P4
12B (M) Since data are supplied and the answer is requested in kilograms (thousands of grams), we can use kilomoles (thousands of moles) to solve the problem. We calculate the amount in kilomoles of POCl 3 that would be produced if each of the reactants were completely converted to product. The smallest of these amounts is the one that is actually produced (this is a limiting reactant question). 1kmol PCl3 10 kmol POCl3 amount POCl3 1.00 kg PCl3 0.0121kmol POCl3 137.33kg PCl3 6 kmol PCl3 10 kmol POCl3 1kmol Cl2 0.0235 kmol POCl3 amount POCl3 1.00 kg Cl2 70.905 kg Cl2 6 kmol Cl2 1kmol P4 O10 10 kmol POCl3 amount POCl3 1.00 kg P4 O10 0.0352 kmol POCl3 283.89 kg P4 O10 1kmol P4 O10
Thus, a maximum of 0.0121 kmol POCl3 can be produced. We next determine the mass of the product. 153.33kg POCl3 mass POCl3 0.0121kmol POCl3 1.86 kg POCl3 1 kmol POCl3
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Chapter 4: Chemical Reactions
13A (M) The 725 g Cl 2 limits the mass of product formed. The P4 s therefore is the reactant in
excess. From the quantity of excess reactant we can find the amount of product formed: 953 g PCl 3 936 g PCl 3 = 17 g PCl 3 . We calculate how much P4 this is, both in the traditional way and by using the initial 215 g P4 and final 953 g PCl3 values of the previous calculation. mass P4 17 g PCl 3
1 mol PCl 3 1 mol P4 123.90 g P4 38 . g P4 137.33g PCl 3 4 mol PCl 3 1 mol P4
13B (M) Find the amount of H2O(l) formed by each reactant, to determine the limiting reactant. 1mol H 2 2 mol H 2 O amount H 2 O 12.2 g H 2 6.05 mol H 2 O 2.016 g H 2 2 mol H 2 1mol O 2 2 mol H 2 O amount H 2 O 154 g O 2 9.63 mol H 2 O 32.00 g O 2 1 mol O 2 Since H 2 is limiting, we must compute the mass of O 2 needed to react with all of the H 2 1mol O 2 32.00 g O 2 mass O 2 reacting = 6.05 mol H 2 O produced 96.8g O 2 reacting 2 mol H 2 O 1mol O 2 mass O 2 remaining = 154 g originally present 96.8g O 2 reacting =57 g O 2 remaining 14A (M) (a) The theoretical yield is the calculated maximum mass of product expected if we were to assume that the reaction has no losses (100% reaction). 1mol CH 2 O 30.03g CH 2 O mass CH 2 O g 1.00 mol CH 3OH 30.0 g CH 2 O 1mol CH 3OH 1mol CH 2 O (b) The actual yield is what is obtained experimentally: 25.7 g CH2O (g). (c) The percent yield is the ratio of actual yield to theoretical yield, multiplied by 100%: 25.7 g CH 2O produced % yield = 100 % = 85.6 % yield 30.0 g CH 2 O calculated 14B (M) First determine the mass of product formed by each reactant. 1mol P4 4 mol PCl3 137.33g PCl3 mass PCl3 = 25.0 g P4 = 111g PCl3 123.90 g P4 1mol P4 1mol PCl3
1mol Cl2 4 mol PCl3 137.33g PCl3 = 118g PCl3 70.91g Cl2 6 mol Cl 2 1mol PCl3 Thus, the limiting reactant is P4 , and 111 g PCl 3 should be produced. This is the theoretical maximum yield. The actual yield is 104 g PCl 3 . Thus, the percent yield of the reaction is 104 g PCl 3 produced 100 % 93.7% yield. 111g PCl 3 calculated mass PCl3 = 91.5g Cl 2
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Chapter 4: Chemical Reactions
15A (M) The reaction is 2 NH 3 (g) + CO 2 (g) CO NH 2 2 (s) + H 2 O(l) . We need to distinguish
between mass of urea produced (actual yield) and mass of urea predicted (theoretical yield). mass CO 2 = 50.0 g CO NH 2 2 produced
44.01g CO 2 1mol CO 2
100.0 g predicted 87.5 g produced
1mol CO NH 2 2
60.1g CO NH 2 2
1mol CO 2
1mol CO NH 2 2
41.8 g CO 2 needed
15B (M) Care must be taken to use the proper units/labels in each conversion factor. Note, you cannot calculate the molar mass of an impure material or mixture. mass C6 H11OH = 45.0 g C 6 H10 produced
100.0 g C6 H10 cal'd
1mol C 6 H10
86.2 g C 6 H10 produc'd 82.1g C 6 H10
1mol C 6 H11OH 1mol C6 H10
100.2 g pure C6 H11OH 100.0 g impure C 6 H11OH 69.0 g impure C 6 H11OH 1mol C 6 H11OH 92.3 g pure C 6 H11OH
16A (M) We can trace the nitrogen through the sequence of reactions. We notice that 4 moles of N (as 4 mol NH 3 ) are consumed in the first reaction, and 4 moles of N (as 4 mol NO) are produced. In the second reaction, 2 moles of N (as 2 mol NO) are consumed and 2 moles of N (as 2 mol NO 2 ) are produced. In the last reaction, 3 moles of N (as 3 mol NO 2 ) are consumed and just 2 moles of N (as 2 mol HNO 3 ) are produced. 1000 g NH 3 1mol NH 3 4 mol NO 2 mol NO 2 mass HNO3 = 1.00 kg NH 3 1kg NH 3 17.03g NH 3 4 mol NH 3 2 mol NO 2 mol HNO 3 63.01g HNO 3 = 2.47 103 g HNO 3 3 mol NO 2 1 mol HNO 3 16B (M) mass KNO3 = 95 g NaN 3
1 mol NaN 3 2 mol KNO3 102 g KNO3 2 mol Na 65.03 g NaN3 2 mol NaN 3 10 mol Na 1 mol KNO3
29.80 30 g KNO3
mass SiO 2 (1) = 1.461 mol NaN 3
2 mol Na 1 mol K 2 O 1 mol SiO 2 64.06 g SiO 2 2 mol NaN 3 10 mol Na 1 mol K 2 O 1 mol SiO 2
9.36 g 9.4 g SiO 2 mass SiO 2 (2) = 1.461 mol NaN 3
2 mol Na 5 mol Na 2 O 1 mol SiO 2 64.06 g SiO 2 2 mol NaN 3 10 mol Na 1 mol Na 2 O 1 mol SiO 2
46.80 g 47 g SiO 2 Therefore, the total mass of SiO2 is the sum of the above two results. Approximately 56 g of SiO2 and 30 g of KNO3 are needed.
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Chapter 4: Chemical Reactions
17A
(D) To determine the mass% for each element, 1 mol Al 3 mol H 2 2.016 g H 2 mass Al = (m) g Al × × × = (m) 0.1121 g Al 26.98 g Al 2 mol Al 1 mol H 2
mass Mg = (1.00-m) g Al ×
1 mol Mg
×
1 mol H 2
24.305 g Mg 1 mol Mg
×
2.016 g H 2 1 mol H 2
= (1.00-m) 0.0829 g Mg Now, we note that the total mass of H2 generated is 0.107 g. Therefore, Mass H2 = (m)(0.1121) + (1.00-m)(0.0829) = 0.107 Solving for m gives a value of 0.82 g. Therefore, mass of Al = 0.83 g. Since the sample is 1.00 g, Mg is 17 wt%.. mass of Mg = 1.00 – 0.83 = 0.17 g, or 17 wt%. 17B (D) Mass of CuO and Cu2O is done in identical fashion to the above problem: mass CuO = (1.500-x) g CuO ×
1 mol CuO
×
1 mol Cu
79.545 g CuO 1 mol CuO
×
63.546 g Cu 1 mol Cu
= (1.500-x) 0.7989 mass Cu 2 O = (x) g Cu 2 O ×
1 mol Cu 2 O
×
2 mol Cu
143.091 g Cu 2 O 1 mol Cu 2 O
×
63.546 g Cu 1 mol Cu
= (x) 0.8882 g Cu 2 O Now, we note that the total mass of pure Cu is 1.2244 g. Therefore,
Mass Cu = (1.500-x)(0.7989) + (x)(0.8882) = 1.2244 Solving for x gives a value of 0.292 g. Therefore, mass of Cu2O = 0.292 mass % of Cu2O = 0.292/1.500 × 100 = 19.47%
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Chapter 4: Chemical Reactions
INTEGRATIVE EXAMPLE A. (D)
Balancing the equation gives the following: C6 H10 O 4 (l) 2NH 3 (g) 4H 2 C6 H16 N 2 (l) 4H 2 O Stepwise approach:
The first step is to calculate the number of moles of each reactant from the masses given. 1000 g 1 mol. 28.4 mol 1 kg 146.16 g 1000 g 1 mol. mol NH 3 = 0.547 kg 32.1 mol 1 kg 17.03 g 1000 g 1 mol. mol H 2 = 0.172 kg 85.3 mol 1 kg 2.016 g To determine the limiting reagent, calculate the number of moles of product that can be obtained from each of the reactants. The reactant yielding the least amount of product is the limiting reagent. mol C6 H10 O 4 = 4.15 kg
mol of C6 H16 N 2 from C6 H10 O4 = 28.4 mol mol of C6 H16 N 2 from NH3 = 32.1 mol
1 mol C6 H16 N 2 28.4 mol 1 mol C6 H10 O 4
1 mol C6 H16 N 2 16.05 mol 2 mol NH 3
1 mol C6 H16 N 2 21.3 mol 4 mol H 2 NH3 yields the fewest moles of product, and is the limiting reagent. mol of C6 H16 N 2 from H 2 = 85.3
To calculate the % yield, the theoretical yield must first be calculated using the limiting reagent: Theoretical yield = 16.05 mol C6 H16 N 2 % yield =
116.22 g C6 H16 N 2 1 kg 1.865 kg 1 mol C6 H16 N 2 1000 g
1.46 kg 78.3% yield 1.865 kg
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Chapter 4: Chemical Reactions
Conversion pathway Approach:
mol of C6 H16 N 2 from C6 H10 O 4 = 4.15 kg C6 H10 O4 mol of C6 H16 N 2 from NH 3 = 0.547 kg NH3
1 mol C6 H16 N 2 1000 g 1 mol. 28.4 mol 1 kg 146.16 g 1 mol C6 H10 O 4
1 mol C6 H16 N 2 1000 g 1 mol. 16.05 mol 1 kg 17.03 g 2 mol NH 3
1 mol C6 H16 N 2 1000 g 1 mol. 21.3 mol 1 kg 2.016 g 4 mol H 2 NH3 yields the fewest moles of product and is therefore the limiting reagent The % yield is determined exactly as above mol of C6 H16 N 2 from H 2 = 0.172 kg H 2
B. (M) Balancing the equation gives the following:
Zn(s) + 2HCl(aq) ZnCl2 (aq) + H 2 (g) Stepwise approach:
To determine the amount of zinc in sample, the amount of HCl reacted has to be calculated first: 1L 0.0134 mol HCl 1000 mL 1L After reaction: 0.0043 M HCl 750.0 mL 0.00323 mol HCl 1000 mL
Before reaction: 0.0179 M HCl 750.0 mL
moles of HCl consumed = 0.0134 – 0.00323 = 0.0102 mol Based on the number of moles of HCl consumed, the number of moles of Zn reacted can be determined: 1 mol Zn 65.39 g Zn 0.3335 g Zn 2 mol HCl 1 mol Zn 0.3335 g Zn reacted Purity of Zn = 100 = 83.4% pure 0.4000 g Zn in sample
0.0102 mol HCl
Conversion pathway Approach:
0.0179 - 0.0043 M HCl
750.0 mL
1L 1000 mL
0.0102 mol HCl
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Chapter 4: Chemical Reactions
Note that we can only subtract concentrations in the above example because the volume has not changed. Had there been a volume change, we would have to individually convert each concentration to moles first. 1 mol Zn 65.39 g Zn 0.0102 mol HCl / 0.4000 g 100 83.4 % Zn 2 mol HCl 1 mol Zn
EXERCISES Writing and Balancing Chemical Equations 1.
(E) (a) 2 SO3 2 SO 2 O 2 (b) Cl2 O7 H 2 O 2 HClO 4 (c) 3 NO 2 H 2 O 2 HNO3 NO (d) PCl3 3 H 2 O H 3 PO3 3 HCl
2.
(E) (a) 3 P2 H 4 4 PH 3 1 2 P4 or 6 P2 H 4 8 PH3 P4 (b) P4 6 Cl2 4 PCl3 (c) 2 FeCl3 3 H 2S Fe 2 S3 6 HCl (d) Mg 3 N 2 6 H 2 O 3 Mg(OH)2 2 NH3
3.
(E) (a) 3 PbO + 2 NH 3 3 Pb + N 2 + 3 H 2 O (b) 2 FeSO4 Fe 2 O3 2 SO 2 +
1
2
O 2 or 4 FeSO 4 2 Fe 2 O3 4 SO 2 + O 2
(c) 6 S2 Cl2 16 NH3 N 4 S4 12 NH 4 Cl +S8 (d) C3 H 7 CHOHCH(C2 H 5 )CH 2 OH 23 2 O 2 8 CO 2 + 9 H 2 O or
4.
2 C3 H 7 CHOHCH(C2 H 5 )CH 2 OH 23 O 2 16 CO 2 +18 H 2 O
(E) (a) SO 2 Cl2 + 8 HI H 2 S + 2 H 2 O + 2 HCl + 4 I 2 (b) FeTiO3 2 H 2SO 4 5 H 2 O FeSO 4 7H 2 O + TiOSO 4 (c) 2 Fe3 O 4 12 HCl +3 Cl2 6 FeCl3 6 H 2 O + O2 (d) C6 H 5 CH 2 SSCH 2 C6 H 5 39 2 O 2 14 CO 2 2 SO 2 7 H 2 O or
2 C6 H 5 CH 2 SSCH 2 C6 H 5 39 O 2 28 CO 2 4 SO 2 14 H 2 O
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Chapter 4: Chemical Reactions
5.
(E) (a) 2 Mg s + O 2 g 2 MgO s (b) 2 NO g + O 2 g 2 NO 2 g (c) 2 C2 H 6 g + 7 O 2 g 4 CO 2 g + 6 H 2 O l (d) Ag 2 SO 4 aq + BaI2 aq BaSO4 s + 2 AgI s
6.
(E) (a) 3Mg s + N 2 g Mg 3 N 2 s (b) KClO3 s KCl s +
3
2
O2 g or
2 KClO3 s 2 KCl s + 3 O2 g
(c) NaOH(s) + NH 4 Cl(s) NaCl(s) + NH 3 (g) + H 2 O g (d) 2 Na(s) + 2H 2 O(l) 2 NaOH(aq) + H 2 (g) 7.
(E) (a) 2 C4 H10 (l) +13O 2 g 8CO 2 g +10 H 2 O l (b) 2 CH 3 CH(OH)CH 3 (l) + 9 O 2 g 6 CO 2 g + 8 H 2 O l (c) CH 3 CH(OH)COOH s + 3O 2 g 3CO 2 g + 3H 2 O l
8.
(E) (a) 2 C3 H 6 g + 9 O 2 g 6 CO 2 g + 6 H 2 O l (b) C6 H 5 COSH s + 9 O 2 g 7 CO 2 g + 3 H 2 O l + SO 2 g (c) 2 CH 2 (OH)CH(OH)CH 2 OH l + 7 O 2 g 6 CO 2 g + 8 H 2 O l
9.
(E) (a) NH 4 NO3 s N2 O g + 2 H2 O g
(b) Na 2 CO3 aq + 2 HCl aq 2 NaCl aq + H 2 O l + CO 2 g (c) 2 CH 4 g + 2 NH 3 g + 3O 2 g 2 HCN g + 6 H 2 O g 10.
(E) (a) 2SO 2 g + O 2 g 2SO3 g (b) CaCO3 s + H 2 O l + CO 2 aq Ca HCO3 2 aq (c) 4 NH 3 g + 6 NO g 5 N 2 g + 6 H 2 O g
11.
(E)
Unbalanced reaction: Balance H atoms: Balance O atoms: Balance N atoms: Multiply by 2 (whole #) Self Check:
N2H4(g) + N2O4(g) H2O(g) + N2(g) N2H4(g) + N2O4(g) 2 H2O(g) + N2(g) N2H4(g) + 1/2 N2O4(g) 2 H2O(g) + N2(g) N2H4(g) + 1/2 N2O4(g) 2 H2O(g) + 3/2 N2(g) 2 N2H4(g) + N2O4(g) 4 H2O(g) + 3 N2(g) 6N+8H+4O 6N+8H+4O
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Chapter 4: Chemical Reactions
12.
(E)
Unbalanced reaction: Balance H atoms: Balance N atoms: Balance O atoms: Multiply by 2 (whole #) Self Check:
NH3(g) + O2(g) 2 NH3(g) + O2(g) 2 NH3(g) + O2(g) 2 NH3(g) + 5/2 O2(g) 4 NH3(g) + 5 O2(g) 4 N + 12 H + 10 O
H2O(g) + NO(g) 3 H2O(g) + NO(g) 3 H2O(g) + 2 NO(g) 3 H2O(g) + 2 NO(g) 6 H2O(g) + 4 NO(g) 4 N + 12 H + 10 O
Stoichiometry of Chemical Reactions 13.
(E) In order to write the balanced chemical equation for the reaction, we will need to determine the formula of the chromium oxide product.
First determine the number of moles of chromium and oxygen, and then calculate the mole ratio. 1 mol Cr 0.01325 mol Cr 52.00 g Cr 1 mol O 2 2 mol O # mol O 0.636 g O 2 0.03975 mol O 32.00 g O 2 1 mol O 2
# mol Cr 0.689 g Cr
0.03975 mol O 3 mol O = Therefore, the formula for the product is CrO3. 0.01325 mol Cr 1 mol Cr Balanced equation =
14.
2 Cr(s) + 3 O2(g) 2 CrO3(s)
(E) In order to write the balanced chemical equation for the reaction, we will need to determine the formula of the manganese oxide product.
First determine the number of moles of manganese and oxygen, and then calculate the mole ratio. # mol O 1.142 g O
1 mol O 0.071375 mol O 16.00 g O
# g Mn = 3.104 g oxide – 1.142 g O = 1.962 g Mn 1 mol Mn # mol Mn 1.962 g Mn 0.03571 mol Mn 54.94 g Mn 0.071375 mol O 2 mol O = Therefore, the formula of the product is MnO2. 0.03571 mol Mn 1 mol Mn Balanced equation =
Mn(s) + O2(g) MnO2(s)
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Chapter 4: Chemical Reactions
15.
(E) The conversion factor is obtained from the balanced chemical equation.
1mol Cl2 7.26 mol Cl2 70.90 g Cl2 2 mol FeCl 3 moles FeCl 3 = 7.26 mol Cl 2 = 4.84 mol FeCl 3 3 mol Cl 2 515 g Cl2
(E) Each calculation uses the stoichiometric coefficients from the balanced chemical equation and the molar mass of the reactant. 1 mol PCl3 Moles PCl3 = 46.3g = 0.337 mol PCl3 137.32 g PCl3 6 mol Cl 2 70.91g Cl 2 mass Cl 2 = 0.337 mol PCl 3 = 35.8 g Cl 2 4 mol PCl 3 1 mol Cl 2 1 mol P4 123.9 g P4 mass P4 = 0.337 mol PCl 3 = 10.4 g P4 4 mol PCl 3 1 mol P4 17. (E) (a) Conversion pathway approach: 1mol KClO3 3mol O2 32.8g KClO3 = 0.401mol O2 122.6 g KClO3 2 mol KClO3 Stepwise approach: 1mol KClO3 32.8g KClO3 = 0.268 mol KClO3 122.6 g KClO3 16.
0.268 mol KClO3 (b)
3mol O 2 = 0.402 mol O 2 2 mol KClO3
Conversion pathway approach: mass KClO3 = 50.0 g O 2
1mol O 2 32.00 g O 2
2 mol KClO3 122.6 g KClO3 128g KClO 3 3 mol O 2 1mol KClO3
Stepwise approach: 50.0 g O 2
1mol O 2 32.00 g O 2
1.56 mol O 2
= 1.56 mol O 2
2 mol KClO3 3 mol O 2
1.04 mol KClO3
= 1.04 mol KClO3
122.6 g KClO3 1mol KClO3
125
= 128 g KClO3
Chapter 4: Chemical Reactions
(c)
Conversion pathway approach:
mass KCl = 28.3g O 2
1mol O 2 2 mol KCl 74.55g KCl = 43.9 g KCl 32.00 g O 2 3mol O 2 1mol KCl
Stepwise approach: 28.3g O 2
1mol O 2 = 0.884 mol O 2 32.00 g O 2
0.884 mol O 2
2 mol KCl = 0.589 mol KCl 3mol O 2
0.589 mol KCl
74.55g KCl = 43.9 g KCl 1mol KCl
1mol Fe 4 mol H 2 2.016 g H 2 2.06 g H 2 55.85g Fe 3mol Fe 1 mol H 2 1 mol Fe 4 mol H 2 O 18.02 g H 2 O (b) mass H 2 O = 63.5 g Fe = 27.3 g H 2 O 55.85 g Fe 3 mol Fe 1 mol H 2 O 1 mol H 2 1mol Fe3O 4 231.54 g Fe3O 4 mass Fe3O 4 = 14.82 g H 2 2.016 g H 2 4 mol H 2 1mol Fe3O 4 (c) = 425g Fe3O 4
18.
(M) (a) mass H 2 = 42.7 g Fe
19.
(M) Balance the given equation, and then solve the problem. 2 Ag 2 CO3 s 4Ag s + 2 CO 2 g + O 2 g mass Ag 2 CO3 = 75.1g Ag
20.
1mol Ag 107.87 g Ag
2 mol Ag 2 CO3 4 mol Ag
275.75 g Ag 2 CO3 1mol Ag 2 CO3
= 96.0 g Ag 2 CO3
(E) The balanced equation is Ca3(PO4)2(s) + 4 HNO3(aq) Ca(H2PO4)2(s) + 2 Ca(NO3)2(aq)
mass HNO3 = 125 kg Ca(H2PO4)2
1 kmol Ca(H 2 PO 4 ) 2 234.05 kg Ca(H 2 PO 4 ) 2
4 kmol HNO 3 1 kmol Ca(H 2 PO 4 ) 2
63.01 kg HNO3 1 kmol HNO3
mass HNO3 = 135 kg HNO3 21.
The balanced equation is CaH2(s) + 2 H2O(l) Ca(OH)2(s) + 2 H2(g) 1mol CaH 2 2 mol H 2 2.016 g H 2 (a) mass H 2 = 127 g CaH 2 = 12.2 g H 2 42.094 g CaH 2 1mol CaH 2 1mol H 2 1mol CaH 2 2 mol H 2 O 18.0153 g H 2 O (b) mass H 2 O = 56.2 g CaH 2 = 48.1 g H 2 O 42.094 g CaH 2 1mol CaH 2 1 mol H 2 O (M)
126
Chapter 4: Chemical Reactions
(c)
mass CaH 2 = 8.12×1024 molecules H 2 ×
1mol H 2 6.022×10
23
molecules H 2
×
1mol CaH 2 2 mol H 2
×
42.094 g CaH 2 1mol CaH 2
mass CaH 2 = 284 g CaH 2
22.
(E) amount O 2 = 156 g CO 2
(b)
mass KO 2 = 100.0 g CO 2
(c)
no. O 2 molecules = 1.00 mg KO 2
23.
1 mol CO 2 3 mol O 2 = 5.32 mol O 2 44.01 g CO 2 2 mol CO 2
(a)
1mol CO 2 44.01g CO 2
4 mol KO 2 2 mol CO 2
71.10 g KO 2 1mol KO 2
= 323.1g KO 2
1 g KO 2 1 mol KO 2 3 mol O 2 1000 mg 71.10 g KO 2 4 mol KO 2
6.022 1023 molecules = 6.35 1018 O 2 molecules 1mol O 2
(M) The balanced equation is Fe 2 O3 s + 3C s 2 Fe l + 3CO g
1 kmol Fe 1 kmol Fe 2 O 3 159.7 kg Fe 2 O 3 = 748 kg Fe 2 O 3 55.85 kg Fe 2 kmol Fe 1 kmol Fe 2 O 3 748 kg Fe 2 O 3 % Fe 2 O 3 in ore = 100% = 79.7% Fe 2 O 3 938 kg ore mass Fe 2 O 3 = 523 kg Fe
24.
heat (M) The following reaction occurs: 2 Ag 2 O s 4 Ag s + O 2 g
mass Ag 2O = 0.187g O 2 % Ag 2 O = 25.
1mol O 2 2 mol Ag 2 O 231.7 g Ag 2 O = 2.71g Ag 2 O 32.0 g O 2 1mol O 2 1mol Ag 2 O
2.71g Ag 2 O 100% = 86.6% Ag 2 O 3.13 g sample
(M) B10H14 + 11 O2 5 B2O3 + 7 H2O
% by mass B10 H14 =
# g B10 H14 100 # g B10 H14 + # g O 2
1 mol B10H14 reacts with 11 mol O2 (exactly) mass B10 H14 = 1 mol B10 H14
mass O 2 = 11 mol O 2
% by mass B10 H14 =
122.22 g B10 H14 = 122.22 g B10 H14 1 mol B10 H14
32.00 g O 2 = 352.00 g O 2 1 mol O 2
122.22 g B10 H14 100 = 25.8% 122.22 g B10 H14 + 352.00 g O 2
127
Chapter 4: Chemical Reactions
26.
(M) 10 Al(s) + 6 NH4ClO4(s) → 4 Al2O3(s) + 2 AlCl3(s) + 12 H2O(l) + 3 N2(g)
1 kg Al
27.
1000 g Al 1 mol Al 6 mol NH 4 ClO 4 117.49 g NH 4ClO 4 2.61103 g NH 4ClO 4 1 kg Al 26.98 g Al 10 mol Al 1 mol NH 4 ClO 4
(E) 2 Al s + 6 HCl aq 2 AlCl3 aq + 3H 2 g . First determine the mass of Al in the foil.
1cm 2.70 g = 9.15g Al 10 mm 1cm3 1 mol Al 3 mol H 2 2.016 g H 2 mass H 2 = 9.15 g Al = 1.03 g H 2 26.98 g Al 2 mol Al 1 mol H 2
mass Al = 10.25cm 5.50 cm 0.601mm
28.
(E) 2 Al s + 6 HCl aq 2 AlCl3 aq + 3 H 2 g
mass H 2 = 225 mL soln
1.088 g 18.0 g HCl 1 mol HCl 3 mol H 2 2.016 g H 2 1 mL 100.0 g soln 36.46 g HCl 6 mol HCl 1 mol H 2
= 1.22 g H 2 29.
(E) First write the balanced chemical equation for each reaction. 2 Na s + 2 HCl aq 2 NaCl aq + H 2 g
2 Al s + 6 HCl aq 2 AlCl3 aq + 3 H 2 g
Mg s + 2 HCl aq MgCl 2 aq + H 2 g Zn s + 2 HCl aq ZnCl 2 aq + H 2 g
Three of the reactions—those of Na, Mg, and Zn—produce 1 mole of H2(g). The one of these three that produces the most hydrogen per gram of metal is the one for which the metal’s atomic mass is the smallest, remembering to compare twice the atomic mass for Na. The atomic masses are: 2 23 u for Na, 24.3 u for Mg, and 65.4 u for Zn. Thus, among these three, Mg produces the most H 2 per gram of metal, specifically 1 mol H 2 per 24.3 g Mg. In the case of Al, 3 moles of H 2 are produced by 2 moles of the metal, or 54 g Al. This reduces as follows: 3 mol H 2 / 54 g Al = 1 mol H 2 / 18 g Al. Thus, Al produces the largest amount of H 2 per gram of metal. 30.
(E) In order for a substance to yield the same mass of CO2(g) per gram of compound as does ethanol when combusted in excess oxygen, the substance must have the same empirical formula. Compound (d), CH3OCH3, is the only compound that fits the description. In fact, compound (d) has the same formula as ethanol, CH3CH2OH, as they are structural isomers, and they should give the same amount of CO2(g) when combusted in excess O2.
128
Chapter 4: Chemical Reactions
Molarity 31.
(M) (a) (b) (c)
32.
2.92 mol CH 3 OH = 0.408 M 7.16 L 7.69 mmol CH 3CH 2OH CH 3CH 2 OH molarity (M) = = 0.154 M 50.00 mL CH 3 OH molarity (M) =
CO NH 2 2 molarity (M) =
25.2 g CO NH 2 2 275 mL
1mol CO NH 2 2
60.06 g CO NH 2 2
1000 mL 1L
= 1.53 M
(E) (a) CH 3CH 2OH molarity (M) = (b)
CH3 2 CO molarity (M) =
(c) C3 H 5 OH 3 molarity (M) = 33.
2.25 104 mol CH 3CH 2OH 1000 mL = 0.00180 M 125 mL 1L
57.5 g CH 3 2 CO 525 mL
1mol CH 3 2 CO
58.08 g CH 3 2 CO
1000 mL 1L
18.5 mL C3 H 5 OH 3 1.26 g 1mol C3 H 5 OH 3 1000 mL × × × = 0.675 M 375 mL soln 1 mL 92.0 9 g C 3 H 5 OH 3 1L
(E) (a) Conversion pathway approach: 150.0 g C12 H 22 O11 1000 mL 1 mol C12 H 22 O11 [C12 H 22 O11 ] = × × = 1.753 M 250.0 mL soln 1L 342.3 g C12 H 22 O11
Stepwise approach: 150.0 g C12 H 22 O11×
1 mol C12 H 22 O11 = 0.4382 mol C12 H 22 O11 342.3 g C12 H 22 O11
1L = 0.2500 L 1000 mL 0.4382 mol C12 H 22 O11 [C12 H 22 O11 ] = = 1.753 M 0.2500 L 250.0 mL soln×
(b)
= 1.886 M
Conversion pathway approach: 98.3 mg solid 97.9 mg CO(NH 2 ) 2 1 mmol CO(NH 2 ) 2 [CO(NH 2 ) 2 ] = × × 5.00 mL soln 100 mg solid 60.06 mg CO(NH 2 ) 2
= 0.320 M CO(NH 2 ) 2
129
Chapter 4: Chemical Reactions
Stepwise approach: 98.3 mg solid ×
97.9 mg CO(NH 2 ) 2 96.2 mg CO(NH 2 ) 2 100 mg solid
96.2 mg CO(NH 2 ) 2 ×
1 g CO(NH 2 ) 2 = 0.0962 g CO(NH 2 ) 2 1000 mg CO(NH 2 ) 2
0.0962 g CO(NH 2 ) 2 ×
1 mol CO(NH 2 ) 2 = 1.60 10-3 mol CO(NH 2 ) 2 60.06 g CO(NH 2 ) 2
1L = 0.00500 L 1000 mL 1.60 10-3 mol CO(NH 2 ) 2 [CO(NH 2 ) 2 ] = = 0.320 M 0.00500 L 5.00 mL soln ×
(c)
Conversion pathway approach: 125.0 mL CH 3 OH 0.792 g 1 mol CH 3 OH [CH 3 OH] = × × = 0.206 M 15.0 L soln 1 mL 32.04 g CH 3 OH Stepwise approach: 125.0 mL CH 3OH 0.792 g 1 mol CH 3OH [CH 3OH] = × × = 0.206 M 15.0 L soln 1 mL 32.04 g CH 3OH 0.792 g = 99.0 g CH 3OH 1 mL 1 mol CH 3OH 99.0 g CH 3OH × = 3.09 mol CH 3OH 32.04 g CH 3OH
125.0 mL CH 3OH ×
[CH 3OH] =
34.
(E) (a) (b) (c)
35.
3.09 mol CH 3OH = 0.206 M 15.0 L soln
0.405g H 2 C4 H 5 NO 4 1000 mL 1mol H 2 C 4 H 5 NO 4 = 0.0304 M 100.0 mL 1L 133.10 g H 2 C4 H 5 NO 4 35.0 mL C3 H 6 O 1000 mL 0.790 g C3 H 6 O 1mol [C3 H 6 O] = = 1.12 M 425 mL soln 1L 1mL 58.08g C3 H 6 O
[H 2 C 4 H 5 NO 4 ] =
[ C 2 H 5 2 O] =
8.8 mg C 2 H 5 2 O 3.00 L soln
1g 1000 mg
1mol C 2 H 5 2 O
74.12 g C 2 H 5 2 O
= 4.0 105 M
(E) (a)
mass C6 H12 O 6 = 75.0 mL soln
(b)
VCH3OH = 2.25 L soln
1L 1000 mL
0.350 mol C6 H12 O 6 1L soln
180.16 g C6 H12 O 6 1mol C6 H12 O 6
= 4.73g
0.485 mol 32.04 g CH 3OH 1mL = 44.1 mL CH 3OH 1L 1mol CH 3OH 0.792 g
130
Chapter 4: Chemical Reactions
36.
(E) (a)
VCH CH OH = 200.0 L soln 3
2
1L 1000 mL
VHCl = 12.0 L 37.
1.65 mol CH 3 CH 2 OH 1L
46.07 g CH 3 CH 2 OH 1mol CH 3CH 2 OH
1mL 0.789 g
= 19.3 L (b)
0.234 mol HCl 36.46 g HCl 100 g soln 1mL soln = 241mL 1L 1mol HCl 36.0 g HCl 1.18g
(M) (a)
85 mg C6 H12 O6 1 mmol C6 H12 O6 1g 10 dL 1 mol C6 H12 O6 1 dL blood 1000 mg 1 L 180.16 g C6 H12 O6 110-3 mol C6 H12 O6
= 4.7
mmol C6 H12 O6 L
(b) Molarity 4.7 103
38. (M) (a) Molarity
mol C6 H12 O6 L
1.2 mg F 1g 1 mol F 5 mol F = 6.3 10 1L 1000 mg 19.00 g F L
6.3 105 mol F1 mol KF 58.1 g KF 1.6 108 L water 5.9 105 g KF (b) # g KF 1 L water 1 mol F 1 mol KF 39.
(E) First we determine each concentration in moles per liter and find the 0.500 M solution. 0.500 g KCl 1 mol KCl 1000 mL = 6.71 M KCl (a) [KCl] = 1 mL 74.551 g KCl 1L 36.0 g KCl 1 mol KCl = 0.483 M KCl (b) [KCl] = 1L 74.551 g KCl 7.46 mg KCl 1 g KCl 1 mol KCl 1000 mL = 0.100 M KCl (c) [KCl] = 1 mL 1000 mg KCl 74.551 g KCl 1L 373 g KCl 1 mol KCl = 0.500 M KCl (d) [KCl] = 10.00 L 74.551 g KCl Solution (d) is a 0.500 M KCl solution.
40.
(E) By inspection, we see that (b) and (c) are the only two that are not per volume of solution. These two solutions need not be considered. A close inspection of the remaining choices reveals that the units for (a) are equivalent to those for (d), that is g NaCl per liter of solution is equivalent to mg NaCl per mL of solution (the mass:volume ratio is the same).
131
Chapter 4: Chemical Reactions
41.
42.
(E) We determine the molar concentration for the 46% by mass sucrose solution. 1 mol C12 H 22 O11 46 g C12 H 22 O11 × 342.3 g C12 H 22 O11 [C12 H 22 O11 ] = = 1.6 M 1 mL 1L 100 g soln× × 1.21 g soln 1000 mL The 46% by mass sucrose solution is the more concentrated. (E) Here we must calculate the CH 3CH 2 OH in the white wine and compare it with
1.71 M CH 3CH 2OH , the concentration of the solution described in Example 4-8. 11 g CH 3CH 2OH 0.95 g soln 1000 mL 1 mol CH 3CH 2 OH CH3CH 2OH = 100.0 g soln 1 mL 1L 46.1 g CH 3CH 2 OH
= 2.3 M CH 3CH 2 OH Thus, the white wine has a greater ethyl alcohol content. 2.05 mol KNO3 1L = 0.0820 M 0.250 L diluted solution
0.01000 L conc'd soln 43.
(E) [KNO3 ] =
44.
(E) Volume of concentrated AgNO3 solution 0.425 mmol AgNO3 1 mL conc. soln. VAgNO3 = 250.0 mL dilute soln × × = 142 mL conc. soln 1 mL dilute soln 0.750 mmol AgNO3
45.
(E) Both the diluted and concentrated solutions contain the same number of moles of K 2SO 4 . This number is given in the numerator of the following expression. 0.198 mol K 2 SO 4 0.125 L 1L K 2 SO 4 molarity = = 0.236 M K2SO4 0.105 L
0.085 mol HCl 1L soln = 1.7 M 0.0250 L
0.500 L dilute sol'n [HCl] =
46.
(E)
47.
(E) Let us compute how many mL of dilute (d) solution we obtain from each mL of concentrated (c) solution. Vc Cc = Vd Cd becomes 1.00 mL 0.250M = x mL 0.0125 M and x = 20 Thus, the ratio of the volume of the volumetric flask to that of the pipet would be 20:1. We could use a 100.0-mL flask and a 5.00-mL pipet, a 1000.0-mL flask and a 50.00-mL pipet, or a 500.0-mL flask and a 25.00-mL pipet. There are many combinations that could be used.
132
Chapter 4: Chemical Reactions
48.
(E) First we must determine the amount of solute in the final solution and then the volume of the initial, more concentrated, solution that must be used. 0.175 mmol KCl 1 mL conc’d soln volume conc’d soln = 250.0 mL = 88.2 mL 1 mL dil soln 0.496 mmol KCl Thus the instructions are as follows: Place 88.2 mL of 0.496 M KCl in a 250-mL volumetric flask. Dilute to the mark with distilled water, stopping to mix thoroughly several times during the addition of water.
Chemical Reactions in Solutions 49.
(M) 1L 0.163 mol AgNO 3 1 mol Na 2S 1000 mL 1 L soln 2 mol AgNO 3 78.05g Na 2S = 0.177 g Na 2 S 1mol Na 2 S
(a) mass Na 2S = 27.8 mL
(b) mass Ag 2S = 0.177 g Na 2S 50.
1mol Na 2S
1mol Ag 2S 247.80 g Ag 2S = 0.562 g Ag 2S 78.05 g Na 2S 1mol Na 2S 1mol Ag 2S
(M) The balanced chemical equation provides us with the conversion factor between the two compounds. (a) mass NaHCO3 = 525 mL solution
0.220 mol Cu NO3 2 2 mol NaHCO3 1L solution 1000 mL solution 1L solution 1mol Cu NO3 2
84.00694 g NaHCO3 = 19.4 g NaHCO3 1 mol NaHCO3 0.220 mol Cu NO3 2 1mol CuCO3 1L solution = 525 mL solution 1000 mL solution 1L 1mol Cu NO3 2
(b) mass CuCO 3
123.6 g CuCO3 = 14.3 g CuCO3 1mol CuCO3
51. (M) The molarity can be expressed as millimoles of solute per milliliter of solution. 0.186 mmol AgNO3 1mmol K 2 CrO 4 1mL K 2 CrO 4 aq VK CrO = 415 mL 2 4 1mL soln 2 mmol AgNO3 0.650 mmol K 2 CrO 4
VK
52.
2CrO4
= 59.4 mL K 2CrO 4 1L 0.477 mol HCl 1mol Ca OH 2 1000 mL 1L soln 2 mol HCl 74.1g Ca OH 2 = 7.33g Ca OH 2 1mol Ca OH 2
(D) (a) mass Ca OH 2 = 415 mL
133
Chapter 4: Chemical Reactions
1.12 kg 24.28 kg HCl 1kmol HCl 1L 100.00 kg soln 36.46 kg HCl 1kmol Ca OH 2 74.10 kg Ca OH 2 = 89.5 kg Ca OH 2 2 kmol HCl 1kmol Ca OH 2
(b) mass Ca OH = 324 L 2
53.
(D) The balanced chemical equation for the reaction is: 2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2H2O(l) # mol HNO3 0.02978 L soln
0.0142 mol Ca(OH) 2 2 mol HNO3 8.46 104 mol HNO3 1 L soln 1 mol Ca(OH) 2
All of the HNO3 that reacts was contained in the initial, undiluted 1.00 mL sample. Since the moles of HNO3 are the same in the diluted and undiluted solutions, one can divide the moles of HNO3 by the volume of the undiluted solution to obtain the molarity. 8.46 104 mol HNO3 mol HNO3 8.46 103 Molarity 0.00100 L L 54.
(M) The balanced chemical equation for the reaction is: H3PO4(aq) + 2 NaOH(aq) Na2HPO4(aq) + 2 H2O(l) 0.217 mol NaOH 1 mol HPO 24 1 1.06 M HPO 24 0.0491 L soln 1 L soln 2 mol NaOH 0.005 L
55.
(M) (a) We know that the Al forms the AlCl 3 . 1mol Al 1mol AlCl3 mol AlCl3 = 1.87 g Al = 0.0693mol AlCl3 26.98g Al 1mol Al 0.0693 mol AlCl3 1000 mL (b) [AlCl3 ] = = 2.91 M AlCl3 23.8 mL 1L
56.
(M) The balanced chemical reaction indicates that 4 mol NaNO 2 are formed from 2 mol Na 2 CO 3 . 138 g Na 2 CO 3 1 mol Na 2 CO 3 4 mol NaNO 2 NaNO 2 = = 1.83 M NaNO 2 1.42 L soln 106.0 g Na 2 CO 3 2 mol Na 2 CO 3
134
Chapter 4: Chemical Reactions
57.
(M) The volume of solution determines the amount of product. 0.186 mol AgNO3 1mol Ag 2 CrO 4 331.73g Ag 2 CrO 4 1L mass Ag 2 CrO 4 = 415 mL 1000 mL 1L soln 2 mol AgNO3 1mol Ag 2 CrO 4 mass Ag 2 CrO 4 = 12.8 g Ag 2 CrO 4
58.
(M) VKMnO4 9.13 g KI
59.
(M)
2 mol KMnO 4 1 L KMnO 4 1 mol KI = 0.138 L KMnO 4 166.0023 g KI 10 mol KI 0.0797 mol KMnO 4
mass Na = 155mL soln
1L 0.175mol NaOH 2 mol Na 22.99g Na 1000 mL 1L soln 2 mol NaOH 1mol Na
= 0.624g Na (M) We determine the amount of HCl present initially, and the amount desired. 1.023 mmol HCl amount HCl present = 250.0 mL = 255.8 mmol HCl 1 mL soln 1.000 mmol HCl amount HCl desired = 250.0 mL = 250.0 mmol HCl 1 mL soln 1mmol Mg 24.3mg Mg mass Mg = 255.8 250.0 mmol HCl = 70. mg Mg 2 mmol HCl 1mmol Mg 61. (M) The mass of oxalic acid enables us to determine the amount of NaOH in the solution. 0.3126 g H 2 C 2O 4 1000 mL 1mol H 2C 2 O 4 2 mol NaOH = 0.2649 M NaOH = 26.21mL soln 1L soln 90.04 g H 2 C2 O 4 1mol H 2 C2 O 4 60.
62.
(D) The total amount of HCl present is the amount that reacted with the CaCO 3 plus the amount that reacted with the Ba(OH)2 (aq). moles HCl from 1mol CaCO3 2 mol HCl 1000 mmol = 0.1000 g CaCO3 CaCO3 reaction 100.09 g CaCO3 1mol CaCO3 1mol = 1.998 mmol HCl moles HCl from 0.01185 mmol Ba OH 2 2 mmol HCl = 43.82 mL Ba OH 2 reaction 1mL soln 1mmol Ba OH 2 = 1.039 mmol HCl The HCl molarity is this total mmol of HCl divided by the total volume of 25.00 mL. 1.998 +1.039 mmol HCl = 0.1215 M [HCl] = 25.00 mL
135
Chapter 4: Chemical Reactions
Determining the Limiting Reactant 63.
(E) The limiting reactant is NH 3 . For every mole of NH3(g) that reacts, a mole of NO(g) forms. Since 3.00 moles of NH3(g) reacts, 3.00 moles of NO(g) forms (1:1 mole ratio).
64.
(E) The reaction of interest is: CaH2(s) + 2 H2O(l) Ca(OH)2(s) + 2 H2(g) The limiting reactant is H2O(l). The mole ratio between water and hydrogen gas is 1:1. Hence, if 1.54 moles of H2O(l) reacts, 1.54 moles of H2(g) forms (1:1 mole ratio).
65.
(M) First we must determine the number of moles of NO produced by each reactant. The one producing the smaller amount of NO is the limiting reactant. 2 mol NO mol NO = 0.696 mol Cu = 0.464 mol NO 3mol Cu Conversion pathway approach: mol NO = 136 mL HNO3 aq ×
6.0 mol HNO3 1L 2 mol NO × × = 0.204 mol NO 1000 mL 1L 8 mol HNO3
Stepwise approach:
1L 0.136 L HNO3 1000 mL 6.0 mol HNO3 0.136 L × 0.816 mol HNO3 1L 2 mol NO 0.816 mol HNO3 × = 0.204 mol NO 8 mol HNO3
136 mL HNO3 aq ×
Since HNO3(aq) is the limiting reactant, it will be completely consumed, leaving some Cu unreacted.
66.
(M) First determine the mass of H 2 produced from each of the reactants. The smaller mass is that produced by the limiting reactant, which is the mass that should be produced. 1mol Al 3mol H 2 2.016g H 2 mass H 2 (Al) = 1.84g Al = 0.206g H 2 26.98g Al 2 mol Al 1mol H 2 2.95mol HCl 3mol H 2 2.016g H 2 1L mass H 2 (HCl) = 75.0 mL = 0.223g H 2 1000 mL 1L 6 mol HCl 1mol H 2 Thus, 0.206 g H 2 should be produced.
136
Chapter 4: Chemical Reactions
67.
(M) First we need to determine the amount of Na 2 CS3 produced from each of the reactants. 2 mol Na 2 CS3 1.26 g 1mol CS2 n Na 2CS3 (from CS2 ) = 92.5 mL CS2 = 1.02 mol Na 2 CS3 1mL 76.14 g CS2 3mol CS2 2 mol Na 2 CS3 n Na 2CS3 (from NaOH) = 2.78 mol NaOH = 0.927 mol Na 2 CS3 6 mol NaOH 154.2 g Na 2 CS3 Thus, the mass produced is 0.927 mol Na 2 CS3 = 143g Na 2 CS3 1mol Na 2 CS3
68.
(D) Since the two reactants combine in an equimolar basis, the one present with the fewer number of moles is the limiting reactant and determines the mass of the products. 0.275 mol ZnSO 4 1L mol ZnSO 4 = 315 mL = 0.0866 mol ZnSO 4 1000 mL 1L soln 1L 0.315 mol BaS mol BaS = 285 mL = 0.0898 mol BaS 1000 mL 1L soln Thus, ZnSO4 is the limiting reactant and 0.0866 mol of each of the products will be produced. 233.4 g BaSO 4 97.46 g ZnS mass products = 0.0866 mol BaSO 4 + 0.0866 mol ZnS 1mol BaSO 4 1mol ZnS = 28.7 g product mixture lithopone
69.
(D) Ca OH 2 s + 2 NH 4 Cl s CaCl2 aq + 2 H 2 O(l) + 2 NH 3 g First compute the amount of NH 3 formed from each reactant in this limiting reactant problem. 2 mol NH 3 1mol NH 4 Cl n NH3 (from NH 4 Cl) = 33.0 g NH 4 Cl = 0.617 mol NH 3 53.49 g NH 4 Cl 2 mol NH 4 Cl 1mol Ca OH 2
2 mol NH 3 = 0.891mol NH 3 74.09 g Ca OH 2 1mol Ca OH 2 Thus, 0.617 mol NH 3 should be produced as NH4Cl is the limiting reagent. n NH3 (from Ca(OH) 2 ) = 33.0 g Ca OH 2
mass NH 3 = 0.617 mol NH 3
17.03g NH 3 1mol NH 3
= 10.5 g NH 3
Now we will determine the mass of reactant in excess, Ca(OH)2. 1mol Ca OH 2 74.09 g Ca OH 2 Ca OH 2 used = 0.617 mol NH3 = 22.9 g Ca OH 2 mol NH3 1mol Ca OH 2
b g
excess mass Ca OH 2 = 33.0 g Ca OH 2 22.9 g Ca OH 2 = 10.1 g excess Ca OH 2
137
2
Chapter 4: Chemical Reactions
70.
(D) The balanced chemical equation is: Ca(OCl)2(s) + 4 HCl(aq) CaCl2(aq) + 2 H2O(l) + 2 Cl2(g) n Cl2 (from Ca(OCl) 2 ) = 50.0 g Ca OCl 2
n Cl2 (from HCl) = 275 mL
1L 1000 mL
1mol Ca OCl 2
142.98 g Ca OCl 2
6.00 mol HCl
1L soln
2 mol Cl 2
1mol Ca OCl 2
2 mol Cl 2 4 mol HCl
= 0.699 mol Cl 2
= 0.825 mol Cl 2
70.91g Cl2 = 49.6 g Cl2 1mol Cl2 The excess reactant is the one that produces the most Cl 2 , namely HCl(aq). The quantity of excess HCl(aq) is determined from the amount of excess Cl2(g) it theoretically could produce (if it were the limiting reagent).
Thus, mass Cl2 expected = 0.699 mol Cl2
4 mol HCl 1000 mL V = 0.825-0.699 mol Cl2 × × excess HCl 2 mol Cl 6.00 mol HCl V = 42.0 mL excess 6.00 M HCl aq excess HCl
mass excess HCl = 0.825 0.699 mol Cl2 71.
2
4mol HCl 36.46g HCl = 9.19 g excess HCl 2mol Cl2 1mol HCl
(M) The number of grams of CrSO4 that can be made from the reaction mixture is determined by finding the limiting reagent, and using the limiting reagent to calculate the mass of product that can be formed. The limiting reagent can determined by calculating the amount of product formed from each of the reactants. Whichever reactant produces the smallest amount of product is the limiting reagent. 2 mol CrSO 4 148.06 g CrSO 4 = 236.90 g CrSO 4 4 mol Zn 1 mol CrSO 4 2 mol CrSO 4 148.06 g CrSO 4 1.7 mol K 2 Cr2 O7 = 503.40 g CrSO 4 1 mol K 2 Cr2 O7 1 mol CrSO 4 3.2 mol Zn
5.0 mol H 2SO 4
2 mol CrSO 4 148.06 g CrSO 4 = 211.51 g CrSO 4 7 mol H 2SO 4 1 mol CrSO 4
H2SO4 is the limiting reagent since it produces the least amount of CrSO4. Therefore, the maximum number of grams of CrSO4 that can be made is 211.51 g.
138
Chapter 4: Chemical Reactions
72.
(M) The number of grams of TiCl4 that can be made from the reaction mixture is determined by finding the limiting reagent, and using the limiting reagent to calculate the mass of product that can be formed. The limiting reagent can determined by calculating the amount of product formed from each of the reactants. Whichever reactant produces the smallest amount of product is the limiting reagent. 1 mol TiO 2 3 mol TiCl4 189.68 g TiCl4 83.1 g TiCl4 79.88 g TiO2 3 mol TiO 2 1 mol TiCl4 1 mol Cl2 3 mol TiCl4 189.68 g TiCl4 45 g Cl2 60.2 g TiCl4 70.90 g Cl2 6 mol Cl2 1 mol TiCl4 1 mol Cl2 3 mol TiCl4 189.68 g TiCl4 130 g TiCl4 11 g C 12.01 g C 4 mol C 1 mol TiCl4 35 g TiO 2
Cl2 is the limiting reagent. Therefore, 6.0×101 g TiCl4 is expected.
Theoretical, Actual, and Percent Yields 73.
(M) 1 mol CCl4 = 1.80 mol CCl4 153.81g CCl4 Since the stoichiometry indicates that 1 mole CCl 2 F2 is produced per mole CCl 4 , the use of 1.80 mole CCl 4 should produce 1.80 mole CCl2 F2 . This is the theoretical yield of the reaction. 1mol CCl2 F2 (b) 187g CCl2 F2 = 1.55 mol CCl2 F2 120.91g CCl2 F2 The actual yield of the reaction is the amount actually produced, 1.55 mol CCl2 F2 . 1.55mol CCl 2 F2 obtained (c) % yield = 100 % 861% . yield 1.80 mol CCl 2 F2 calculated
(a)
74.
277 grams CCl4
(M) (a)
mass C 6 H 10 = 100.0 g C 6 H 11OH
1 mol C 6 H 11OH 1 mol C 6 H 10 100.16 g C 6 H 11OH 1 mol C 6 H 11OH
82.146g C6 H10 = 82.01g C6 H10 = theoretical yield 1mol C6 H10 64.0g C 6 H 10 produced (b) percent yield = 100 % 78.0% yield 82.01g C 6 H 10 calculated 1.000 g calculated 1 mol C 6 H 10 (c) mass C 6 H11OH = 100.0 g C6 H10 produced 0.780 g produced 82.15 g C 6 H 10 1mol C6 H11OH 100.2 g C6 H11OH = 156 g C6 H11OH are needed 1mol C6 H10 1mol C6 H11OH
139
Chapter 4: Chemical Reactions
75.
actual yield 100 % theoretical yield The actual yield is given in the problem and is equal to 28.2 g. In order to determine the theoretical yield, we must find the limiting reagent and do stoichiometry.
(D) % yield =
Conversion pathway approach: 1 mol Al 2 O3 2 mol Na 3 AlF6 209.94 g Na 3 AlF6 32.2 g Na 3 AlF6 7.81 g Al 2 O3 101.96 g Al 2 O3 1 mol Al 2 O3 1 mol Na 3 AlF6 0.141 mol 2 mol Na 3 AlF6 209.94 g Na 3 AlF6 34.5 g Na 3 AlF6 3.50 L 1L 6 mol NaOH 1 mol Na 3 AlF6 Stepwise approach: Amount of Na3AlF6 produced from Al2O3 if all Al2O3 reacts 1 mol Al 2 O3 7.81 g Al 2 O3 = 0.0766 mol Al 2 O3 101.96 g Al 2 O3
0.0766 mol Al 2 O3
2 mol Na 3 AlF6 = 0.153 mol Na 3 AlF6 1 mol Al 2 O3
209.94 g Na 3 AlF6 32.1 g Na 3AlF6 1 mol Na 3 AlF6 Amount of Na3AlF6 produced from NaOH if all NaOH reacts 0.141 mol NaOH 0.494 mol NaOH 3.50 L 1L 2 mol Na 3 AlF6 0.494 mol NaOH = 0.165 mol Na 3 AlF6 6 mol NaOH 209.94 g Na 3 AlF6 0.165 mol Na 3 AlF6 34.5 g Na 3AlF6 1 mol Na 3 AlF6 0.153 mol Na 3AlF6
Al2O3 is the limiting reagent. 76.
% yield =
28.2 g 100 % 87.6% 32.2 g
(D) The balanced equation is 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(g) 6.63 g 100 % theoretical yield 1 mol NH 3 1 mol N 2 28.01 g N 2 14.9 g N 2 18.1 g NH 3 17.03 g NH 3 2 mol NH 3 1 mol N 2 % yield =
140
Chapter 4: Chemical Reactions
1 mol N 2 28.01 g N 2 1 mol CuO 10.6 g N 2 79.55 g CuO 3 mol CuO 1 mol N 2 CuO is the limiting reagent. The theoretical yield of N2 is 10.6 grams. 90.4 g CuO
% yield =
6.63 g 100 % 62.5 % 10.6 g
77.
(M) (a) We first need to solve the limiting reactant problem involved here. 1mol C4 H9 OH 1mol C4 H9 Br n C4H9Br (from C 4 H 9 OH) = 15.0 g C4 H9 OH = 0.202 mol C4 H9 Br 74.12 g C4 H 9 OH 1mol C4 H 9 OH 1mol NaBr 1mol C4 H 9 Br n C4 H9 Br (from NaBr) = 22.4 g NaBr = 0.218 mol C4 H 9 Br 102.9 g NaBr 1mol NaBr 1mol H 2 SO 4 1mol C4 H 9 Br n C4 H9 Br (from H 2 SO 4 ) = 32.7 g H 2 SO 4 = 0.333 mol C 4 H 9 Br 98.1g H 2 SO 4 1mol H 2 SO 4 137.0 g C4 H9 Br Thus the theoretical yield of C4 H9 Br = 0.202 mol C4 H9 Br = 27.7 g C4 H9 Br 1mol C4 H9 Br (b) The actual yield is the mass obtained, 17.1 g C4 H 9 Br . 17.1g C 4 H 9 Br produced (c) Then, % yield = 100% = 61.7% yield 27.7 g C 4 H 9 Br expected
78.
(M) (a) Again, we solve the limiting reactant problem first. 1000 mL 1.20 g 1mol C6 H 5 NO 2 n C6 H5 N (from C 6 H 5 NO 2 ) = 0.10 L C6 H 5 NO 2 2 1L 1mL 123.1g C6 H 5 NO 2
n C6 H 5 N
2
(from C 6 H14 O 4 )
1mol C6 H 5 N 2
= 0.49 mol C6 H 5 N 2
2 mol C6 H 5 NO 2
= 0.30 L C6 H14 O 4
1mol C6 H 5 N 2 4 mol C6 H14 O 4
1000 mL 1.12 g 1mol C6 H14 O 4 1L 1mL 150.2 g C6 H14 O 4
= 0.56 mol C6 H 5 N 2
Thus the theoretical yield for C6 H 5 N 2 = 0.49 mol C 6 H 5 N 2
(b)
actual yield = 55 g C6 H 5 N 2 produced
(c)
percent yield =
55g C6 H 5 N 2 produced 89 g C6 H 5 N 2 expected
141
182.2 g C 6 H 5 N 2 1mol C 6 H 5 N 2
100% 62% yield
= 89 g C 6 H 5 N 2
Chapter 4: Chemical Reactions
79.
(M) Balanced equation: 3CH 3COOH + PCl3 3CH 3COCl + H 3 PO3 100.0 g calculated 1mol CH 3COCl 3mol CH 3COOH mass acid = 75g CH 3COCl 78.2 g produced 78.5g CH3COCl 3mol CH3COCl 60.1g pure CH 3COOH 100 g commercial = 76 g commercial CH3COOH 1mol CH 3COOH 97 g pure CH 3COOH
80.
(M) mass CH 2 Cl 2 = 112 g CH 4
81.
(E) A less-than-100% yield of desired product in synthesis reactions is always the case. This is because of side reactions that yield products other than those desired and because of the loss of material on the glassware, on filter paper, etc. during the various steps of the procedure. A main criterion for choosing a synthesis reaction is how economically it can be run. In the analysis of a compound, on the other hand, it is essential that all of the material present be detected. Therefore, a 100% yield is required; none of the material present in the sample can be lost during the analysis. Therefore analysis reactions are carefully chosen to meet this 100 % yield criterion; they need not be economical to run.
82.
(M) The theoretical yield is 2.07 g Ag 2 CrO 4 . If the mass actually obtained is less than this, it is likely that some of the pure material was not recovered, perhaps stuck to the walls of the flask in which the reaction occurred, or left suspended in the solution. Thus while it is almost a certainty that less than 2.07 g will be obtained, the absolute maximum mass of Ag2CrO4 expected is 2.07 g. If the precipitate weighs more than 2.07 g, the extra mass must be impurities (e.g., the precipitate was not thoroughly dried).
1 mol CH 4 1 mol CH 3Cl 0.92 mol CH 3Cl produced 16.04 g CH 4 1 mol CH 4 1.00 mol CH 3Cl expected 1mol CH 2 Cl2 84.93g CH 2 Cl2 0.92 g CH 2 Cl2 produced = 5.0 102 g CH 2 Cl2 1mol CH 3Cl 1mol CH 2 Cl2 1.00 g CH 2 Cl 2 calculated
Consecutive Reactions, Simultaneous Reactions 83.
(D) We must determine the amount of HCl needed to react with each component of the mixture. Mg OH 2 (s) + 2 HCl(aq) MgCl 2 (aq) + 2 H 2O(l) MgCl 2 (aq) + H 2O(l) + CO 2 (g) 35.2 g MgCO3 1mol MgCO3 2 mol HCl n HCl (consumed by MgCO 3 ) = 425g mixt. = 3.55 mol HCl 100.0 g mixt. 84.3g MgCO3 1mol MgCO3 MgCO3 (s) + 2 HCl(aq)
n HCl (consumed by Mg(OH) 2 ) = 425 g mixt.
64.8 g Mg OH 2 100.0 g mixt.
mass HCl = 3.55 + 9.45 mol HCl
1mol Mg OH 2
58.3 g Mg OH 2
36.46 g HCl = 474 g HCl 1mol HCl
142
2 mol HCl 1mol MgCO 3
= 9.45 mol HCl
Chapter 4: Chemical Reactions
84. (D) Here we need to determine the amount of CO 2 produced from each reactant. C3 H8 g + 5O 2 g 3CO 2 g + 4 H 2 O l
2 C4 H10 g +13O 2 g 8CO 2 g +10 H 2 O l 72.7g C3 H8 1mol C3 H8 3mol CO 2 = 20.1mol CO 2 100.0 g mixt. 44.10 g C3 H8 1mol C3 H8 27.3g C4 H10 1mol C4 H10 8 mol CO 2 n CO2 (from C 4 H10 ) = 406 g mixt. = 7.63mol CO 2 100.0 g mixt 58.12 g C4 H10 2 mol C4 H10 44.01g CO 2 mass CO 2 = 20.1 + 7.63 mol CO 2 = 1.22 103 g CO 2 1mol CO 2
n CO2 (from C3 H 8 ) = 406 g mixt.
85. (M) The molar ratios given by the stoichiometric coefficients in the balanced chemical equations are used in the solution, namely CH4(g) + 4 Cl2(g) CCl4(g) + 4 HCl(g) CCl4(g) + 2 HF (g) CCl2F2(g) + 2 HCl(g) 1mol CCl 2 F2 1mol CCl 4 4 mol Cl 2 amount Cl2 = 2.25 103 g CCl2 F2 = 74.4 mol Cl2 120.91g CCl2 F2 1mol CCl2 F2 1mol CCl4 86. (M) Balanced Equations: C2H6(g) + 7 2 O2(g) 2 CO2(g) + 3 H2O(l) CO2(g) + Ba(OH)2(aq) BaCO3(s) + H2O(l) Conversion pathway approach: mass C2 H6 = 0.506 g BaCO3
1mol BaCO3 197.3g BaCO3
1mol CO 2 1mol BaCO3
Stepwise approach: 0.506 g BaCO 3
1mol BaCO3 197.3g BaCO3
2.56 10-3mol BaCO3 2.56 10-3mol CO 2
= 2.56 10-3 mol BaCO3
1mol CO 2 1mol BaCO3
2 mol C 2 H 6 4 mol CO 2
1.28 10-3 mol C 2 H 6
= 2.56 10-3 mol CO 2
= 1.28 10-3 mol C2 H 6
30.07 g C 2 H 6 = 0.0386 g C 2 H 6 1 mol C 2 H 6
143
2 mol C 2 H 6 4 mol CO 2
30.07 g C 2 H 6 1mol C 2 H 6
= 0.0386 g C 2 H 6
Chapter 4: Chemical Reactions
87. (D) NaI(aq)+ AgNO3(aq) AgI(s )+ NaNO3(aq) (multiply by 4) (multiply by 2) 2 AgI(s) + Fe(s) FeI2(aq) + 2 Ag(s) (unchanged) 2 FeI2(aq) + 3 Cl2(g) 2 FeCl3(aq) + 2 I2(s) 4NaI(aq) + 4AgNO3(aq) + 2Fe(s) + 3Cl2(g) 4NaNO3(aq) + 4Ag(s) + 2FeCl3(aq) + 2I2(s)
For every 4 moles of AgNO3, 2 moles of I2(s) are produced. The mass of AgNO3 required 1 mol I 2 (s) 4 mol AgNO3 (s) 169.873 g AgNO3 (s) 1000 g I 2 (s) = 1.00 kg I2(s) 1 kg I 2 (s) 253.809 g I 2 (s) 2 mol I 2 (s) 1 mol AgNO3 (s) = 1338.59 g AgNO3 per kg of I2 produced or 1.34 kg AgNO3 per kg of I2 produced 88. (D) Fe + Br2 FeBr2 (multiply by 3) Fe3Br8 3 FeBr2 + Br2 Fe3Br8 + 4 Na2CO3 8 NaBr + 4 CO2 + Fe3O4 3 Fe + 4 Br2 + 4 Na2CO3 8 NaBr + 4 CO2 + Fe3O4 Hence, 3 moles Fe(s) forms 8 mol NaBr
MassFe consumed = 2.50103 kg NaBr = 509103 g Fe
1000 g NaBr 1 kg NaBr
1 mol NaBr 102.894 g NaBr
3 mol Fe 8 mol NaBr
55.847 g Fe 1 mol Fe
1 kg Fe = 509 kg Fe required to produce 2.5 103 kg KBr 1000 g Fe
89. (M) (a) Si(s) + 2 CO(g) SiO2(s) + 2 C(s) Si(s) + 2 Cl2(g) → SiCl4(l) SiCl4(l) + 2 H2(g) → Si(s, ultrapure) + 4 HCl(g) (b) 1 kg Si (ultrapure, s)
1 mol SiCl 4 1000 g 1 mol Si 1 mol Si 1kg 28.09 g 1 mol Si ultrapure 1 mol SiCl4
2 mol C 12.01 g C = 885 g C 1 mol Si 1 mol C 1 mol SiCl 4 2 mol Cl2 1000 g 1 mol Si 1 kg Si (ultrapure, s) 1kg 28.09 g 1 mol Si ultrapure 1 mol SiCl4
70.91 g Cl2 = 5.05 103 g Cl 2 1 mol Cl2
1 kg Si (ultrapure, s)
1000 g 1 mol Si 2 mol H 2 2.016 g H 2 = 144 g H 2 1kg 28.09 g 1 mol Si ultrapure 1 mol H 2
144
Chapter 4: Chemical Reactions
90. (D)
1 kg HNO3
1 mol HNO3 3 mol NO 2 1000 g 2 mol NO 1 kg 63.02 g HNO3 2 mol HNO3 2 mol NO 2
4 mol NH 3 3 mol H 2 2.016 g H 2 = 71.98 g H 2 4 mol NO 2 mol NH3 1mol 1 mol HNO3 1000 g 3 mol NO 2 2 mol NO 1 kg HNO3 1 kg 63.02 g HNO3 2 mol HNO3 2 mol NO 2 4 mol NH 3 1 mol N 2 28.02 g N 2 = 333.5 g N 2 4 mol NO 2 mol NH3 1mol In order to determine the mole ratio between O2 and HNO3, it is helpful to sum the individual steps and write the overall reaction, since O2 is listed as a reactant in two different steps. The four reactions cannot simply be added together. Rather, we must figure out how to add all four steps so that N2, H2 an O2 are the only reactants, and NO is not an intermediate but is one of the final products. First, multiply step 1 by a factor of two and add it to step 2: 2 × [ N2(g) + 3 H2(g) → 2 NH3(g) ] 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The resulting reaction is: 2N2 + 6H2 + 5O2 → 6 H2O + 4 NO Which can be summed with step 3 multiplied by a factor of two: 2 × [2 NO + O2 → 2NO2] The resulting reaction is: 2 N2 + 6 H2 + 7 O2 → 6 H2O + 4 NO2 Multiply the resulting reaction by a factor of three: 3 × [2 N2 + 6 H2 + 7 O2 → 6 H2O + 4 NO2 ] Sum the above reaction with step 4 multiplied by a factor of four: 4 × [3 NO2 + H2O → 2 HNO3 + NO] The overall reaction is therefore: 6 N2 + 18 H2 + 21 O2 1 kg HNO3
8 HNO3 + 4 NO + 14 H2O
1 mol HNO3 21 mol O 2 31.995 g O 2 1000 g = 1333 g O 2 1 kg 63.02 g HNO3 8 mol HNO3 1mol
Alternatively, another method of solving for the number of grams of O2 is: number of mol HNO3 = 1000 g / 63.02 g mol−1 = 15.87 mol
15.87 mol HNO3
3 mol NO 2 5 mol O 2 32.00 g O 2 2 mol NO 1 mol O 2 1.33 103 g O 2 2 mol HNO3 2 mol NO 2 2 mol NO 4 mol NO 1 mol O 2
145
Chapter 4: Chemical Reactions
91. (D) The reactions are as follows. MgCO3(s) → MgO(s) + CO2(g) CaCO3(s) → CaO(s) + CO2(g)
% by mass of MgCO3 =
g MgCO3 100 % g MgCO3 + g CaCO3
Let m = mass, in grams, of MgCO3 in the mixture and let 24.00 − m = mass in grams of CaCO3 in the mixture. Convert from g MgCO3 to g CO2 to obtain an expression for the mass of CO2 produced by the first reaction. 1 mol MgCO3 1 mol CO 2 44.01 g CO 2 g CO 2 from MgCO3 m g MgCO3 84.32 g MgCO3 1 mol MgCO3 1 mol CO 2 Convert from g CaCO3 to g CO2 to obtain an expression for the mass of CO2 produced by the second reaction. 1 mol CaCO3 1 mol CO 2 44.01 g CO 2 g CO 2 from CaCO3 (24.00 m) g CaCO3 100.09 g CaCO3 1 mol CaCO3 1 mol CO 2 The sum of these two expressions is equal to 12.00 g CO2. Thus: 44.01 44.01 m 84.32 (24.00 m) 100.09 12.00 Solve for m: m = 17.60 g % by mass of MgCO3 =
17.60 g 100 % = 73.33 % 24.00 g
92. (D) Assuming the mass of the sample is 100.0 g, the sample contains 72.0 g Fe.
% by mass of Fe 2 O3 =
g Fe 2 O3 100 % g Fe 2 O3 + g FeO
Total mass of Fe in sample = mass of Fe from Fe2O3 + mass of Fe from FeO Let m be the mass, in grams, of Fe2O3 in the mixture. 1 mol Fe 2 O3 2 mol Fe 55.85 g Fe g Fe from Fe 2 O3 m g Fe 2 O3 159.7 g Fe 2 O3 1 mol Fe 2 O3 1 mol Fe The mass of FeO is (100 − m) grams.
146
Chapter 4: Chemical Reactions
g Fe from FeO (100 m) g FeO
1 mol FeO 1 mol Fe 55.85 g Fe 71.85 g FeO 1 mol FeO 1 mol Fe
The sum of these two expressions is equal to 72.0 g Fe. Thus: 55.85 55.85 72.0 = m 2 (100.0 m) 159.7 71.85 Solve for m:
m
% by mass of Fe 2O3 =
=
73.6 g
73.6 g 100 % = 73.6% 100.0 g
Integrative and Advanced Exercises 93. (E) CaO(s) CO 2 (g) (a) CaCO 3 (s) (b) 2 ZnS(s) 3 O 2 (g) 2 ZnO(s) 2 SO 2 (g) (c) C 3 H 8 (g) 3 H 2 O(g) 3 CO(g) 7 H 2 (g)
(d) 4 SO2(g) + 2 Na2S(aq) + Na2CO3(aq) CO2(g) + 3 Na2S2O3(aq) 94. (E) (a) 4 Ca 3 (PO 4 )2 12 SiO 2 (s) + 20 C(s) 12 CaSiO3 (s) 2 P4 (g) 20 CO(g) P4 (s) + 6 Cl 2 (g) 4 PCl3 (g) PCl3 (g) +3 H 2 O(l) H 3 PO3 (aq) + 3 HCl(aq)
(b) 2 Cu(s) O 2 (g) CO 2 (g) H 2 O(g) Cu 2 (OH) 2 CO 3 (s) 6 H2O (c) P4(s) + 5 O2(g) P4O10(s) 4 H3PO4(aq)
(d) 3Ca(H 2 PO 4 ) 2 (aq) 8 NaHCO3 (aq) Ca 3 (PO 4 ) 2 (aq) 4 Na 2 HPO 4 (aq) 8 CO 2 (g) 8 H 2 O(l)
95. (M) The balanced equation is as follows:
2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Conversion pathway approach: g LiOH
1.00 103 g CO 2 1 mol CO 2 2 mol LiOH 23.95 g LiOH 3 astronauts 6 days astronaut day 44.01 g CO2 1 mol CO2 1 mol LiOH
1.96 104 g LiOH
147
Chapter 4: Chemical Reactions
Stepwise approach: 1.00 103 g CO 2 1 mol CO 2 mol CO 2 = 22.7 astronaut day 44.01 g CO 2 astronaut day mol CO 2 mol CO 2 22.7 3 astronauts 68.2 astronaut day day mol CO 2 68.2 6 days 409 mol CO 2 day 2 mol LiOH 409 mol CO 2 = 818 mol LiOH 1 mol CO 2 23.95 g LiOH 818 mol LiOH 1.96 104 g LiOH 1 mol LiOH 96. (M) mass CaCO3 = 0.981g CO 2
% CaCO 3 =
1mol CO 2 44.01g CO 2
1mol CaCO3 100.1g CaCO3 = 2.23g CaCO3 1mol CO 2 1mol CaCO3
2.23 g CaCO 3 100% = 68.0% CaCO 3 (by mass) 3.28 g sample
97. (D) We determine the empirical formula, basing our calculation on 100.0 g of the compound. 1 mol Fe amount Fe 72.3 g Fe 1.29 mol Fe 1.29 1.00 mol Fe 55.85 g Fe 1 mol O amount O 27.7 g O 1.73 mol O 1.29 1.34 mol O 16.00 g O The empirical formula is Fe 3 O 4 and the balanced equation is as follows. 3Fe 2 O3 (s) + H 2 (g) 2 Fe3 O 4 (s) + H 2 O(g) 98. (D) Assume 100g of the compound FexSy, then: Number of moles of S atoms = 36.5g/32.066 g S/mol = 1.138 moles Number of moles of Fe atoms = 63.5g/ 55.847g Fe/mol = 1.137 moles So the empirical formula for the iron –containing reactant is FeS Assume 100g of the compound FexOy, then: Number of moles of O atoms = 27.6g/16.0 g O/mol = 1.725 moles Number of moles of Fe atoms = 72.4g/ 55.847g Fe/mol = 1.296 moles So the empirical formula for the iron-containing product is Fe3O4 Balanced equation: 3 FeS + 5 O2 Fe3O4 + 3 SO2
148
Chapter 4: Chemical Reactions
99.
(M)
M CH 3CH 2OH =
mol CH 3CH 2 OH volume of solution
mol CH 3CH 2OH 50.0 mL 0.7893
Molarity
g CH 3CH 2 OH 1 mol CH 3CH 2 OH = 0.857 mol mL 46.07 g CH 3CH 2 OH
0.857 mol CH 3CH 2OH 8.88 M CH 3CH 2 OH 0.0965 L solution
100. (M) (a) H2O volume = 72.061 g × 1 mL / 0.99705 g = 72.274 mL. CH3OH volume = 192.25 g × 1 mL / 0.78706 g = 244.26 mL. The volume sum of the two pure liquids is 316.53 mL.
Masses are always additive. Mass of solution = 72.061 g + 192.25 g = 264.31 g. Volume of solution = 264.31 g × 1 mL / 0.86070 g = 307.09 mL. The volume of the solution is 9.44 mL less than the sum of the pure liquid volumes. The volumes are not additive. 1 mol CH 3OH 32.0422 g CH 3OH mol CH 3OH 19.538 0.30709 L L
192.25 g CH 3OH (b)
101.
Molarity
(D) Let V be the volume of 0.149 M HCl(aq) that is required.
moles of HCl in solution C = moles HCl in solution A + moles HCl in solution B (V + 0.100) × 0.205 M = (V × 0.149 M ) + (0.100 × 0.285 M) Solve for V:
V = 0.143 L = 143 mL
102. (D) Let V be the volume of 0.0175 M CH3OH(aq) that is required.
moles CH3OH in solution C (V + 0.0500) × 0.0200 M
= moles CH3OH in solution A + moles CH3OH in solution B = (V × 0.0175 M) + (0.050 × 0.0248 M)
Solve for V: V = 0.0960 L = 96.0 mL
149
Chapter 4: Chemical Reactions
103. (M) 1.52 g Na × 1 g sol × 1000 mL × 1 mol Na × 1 mol NaCl × 58.4425 g NaCl 1×106 g sol 1 mL sol 1L 22.9898 g Na 1 mol Na 1 mol NaCl = 0.003864 M NaCl 104.
(D) mass Ca(NO 3 ) 2 50.0 L soln
1000 mL 1.00 g soln 2.35 g Ca 1 mol Ca 1L 1 mL soln 10 6 g soln 40.08 g Ca
1 mol Ca(NO 3 ) 2 164.09 g Ca(NO 3 ) 2 1000 mg 481 mg Ca(NO 3 ) 2 1 mol Ca 1 mol Ca(NO 3 ) 2 1g
105. (D) We can compute the volume of Al that reacts with the given quantity of HCl. 1L 12.0 mol HCl 2 mol Al 27.0 g Al 1cm3 VAl 0.05 mL 0.002 cm3 1000 mL 1L 6 mol HCl 1mol Al 2.70 g Al
0.002 cm3 10 mm volume area 0.2 cm 2 thickness 0.10 mm 1cm 106.
(D) Here we need to determine the amount of HCl before and after reaction; the difference is the amount of HCl that reacted. 1.035 mol HCl initial amount HCl 0.05000 L 0.05175 mol HCl 1L 0.812 mol HCl final amount HCl 0.05000 L 0.0406 mol HCl 1L 1 mol Zn 65.39 g Zn mass Zn (0.05175 0.0406) mol HCl 0.365 g Zn 2 mol HCl 1 mol Zn
107.
Let us first determine the moles of NH4NO3 in the dilute solution. 2.37×10-3 g N 1 mol N 1 mol NH 4 NO3 mass NH4 NO3 = 1000 mL× × × = 0.0846 mol NH 4 NO3 1 mL 14.007 g N 2 mol N 1 L soln 1000 mL 118 mL soln volume of solution 0.0846 mol NH 4 NO 3 0.715 mol NH 4 NO 3 1L (M)
108. (D) First we determine the molarity of seawater. 2.8 g NaCl 1.03 g 1000 mL 1 mol NaCl 0.49 M NaCl concentration 100.0 g soln 1 mL soln 1 L soln 58.44 g NaCl
0.49 M 1.00 10 6 L 9.0 10 4 L 5.45 M 6 4 5 Volume to be evaporated 1.00 10 L 9.0 10 L 9.1 10 L of water to be evaporated Then the volume of the final solution: c1V1 c 2V2
150
V2
Chapter 4: Chemical Reactions
109. (D) Here we must determine the amount of PbI 2 produced from each solute in this limiting reactant problem. 1mol PbI 2 1.093g 0.120 g KI 1mol KI n PbI2 (from KI) 99.8 mL 0.0394 mol 1mL soln 1g soln 166.00 g KI 2 mol KI 1mol PbI 2 1.134 g 0.140 g Pb(NO3 ) 2 1mol Pb(NO3 ) 2 n PbI2 (from Pb(NO3 ) 2 ) 96.7 mL 1mL soln 1g soln 331.2 g 1mol Pb(NO3 ) 2 0.0464 mol PbI 2 Then the mass of PbI 2 is computed from the smaller amount produced. 461.0 g PbI 2 mass PbI 2 0.0394 mol PbI 2 18.2 g PbI 2 1 mol PbI 2
CaCO 3 (s) 2 HCl(aq) CaCl 2 (aq) H 2 O CO 2 (g) 1 mol CaCO 3 2 mol HCl Amount of HCl reacted 45.0 g CaCO 3 0.899 mol HCl 100.09 g CaCO 3 1 mol CaCO 3 1000 mL 1.13 g 0.257g HCl 1 mol HCl Init. amt. of HCl 1.25 L soln 9.96 mol HCl 1L 1 mL 1 g soln 36.46 g HCl 9.96 mol HCl 0.899 mol HCl Final HCl concentration 7.25 M HCl 1.25 L soln
110. (D)
111. (D) We allow the mass of Al in the alloy to be represented by x. We then set up an expression for determining the mass of H2 and solve this expression for x. 1 mol Fe 1 mol H 2 2.02 g H 2 0.105 g H 2 (2.05 x ) g Fe 55.85 g Fe 1 mol Fe 1 mol H 2
1 mol Al 3 mol H 2 2.02 g H 2 x g Al 27.0 g Al 2 mol Al 1 mol H 2 0.105 g H 2 (2.05 x)0.0362 0.112 x 0.0742 (0.112 0.0362) x 0.0742 0.076 x
0.076 x 0.105 0.0742 0.031 %Al
x
0.41 g Al 20.% Al (by mass) 2.05 g alloy
151
0.031 0.41 g Al 0.076 and the alloy is also 80. % Fe (by mass).
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112. (D)
Mg(s) 2 HCl(aq) MgCl2 (aq) H 2 (g) 2 Al(s) 6 HCl(aq) 2 AlCl 3 (aq) 3 H 2 (g)
Let x represent the mass of Mg. 0.0163 g H 2
1 mol H 2 2.016 g H 2
x g Mg
1 mol Mg
0.0367 g Mg 0.155 g alloy
1 mol H 2
24.305 g Mg 1 mol Mg
0.00809 0.041144 x 0.00862 0.055593 x % Mg
x
(0.155 x)g Al
0.00862 0.00809 0.055593 0.041144
1 mol Al 26.982 g Al
3 mol H 2 2 mol Al
0.0367 g Mg
100% 23.7% Mg or~24% Mg (by mass).
113. (D) One way to solve this problem would be to calculate the mass of CO2 produced from a 0.220 g sample of each alcohol. The results are 0.303 g CO2 from 0.220 g CH3OH and 0.421 g CO2 from 0.220 g CH3CH2OH. Obviously a mixture has been burned. But we have sufficient information to determine the composition of the mixture. First we need the balanced equations for the combustion reactions. Then, we represent the mass of CH3OH by x. 2 CH 3OH(l) 3 O 2 (g) 2 CO 2 (g) 4 H 2O(l)
CH 3CH 2 OH(l) 3 O 2 (g) 2 CO 2 (g) 3 H 2O(l)
1 mol CH 3OH 2 mol CO 2 44.01 g CO 2 mass CO 2 0.352 g CO 2 x g CH 3OH 32.04 g CH 3OH 2 mol CH 3OH 1 mol CO 2 1 mol CH 3CH 2 OH 2 mol CO 2 44.01 g CO 2 (0.220 x) g CH 3CH 2 OH 46.07 g CH 3CH 2OH 1 mol C 2 H 5OH 1 mol CO 2 1.374 x (0.220 x)1.911 0.421 0.537 x 0.068 0.127 g CH 3OH 0.537 By difference, the mass of CH3CH 2 OH is 0.220 g 0.127 g 0.093 g CH 3OH 0.537 x 0.352 0.420 0.068 or x
114. (D) CH 3CH 2OH(l) 3O 2 (g) 2 CO 2 (g) 3 H 2 O(l)
(CH 3CH 2 ) 2 O (l) 6 O 2 (g) 4 CO 2 (g) 5 H 2O(l) Since this is classic mixture problem, we can use the systems of equations method to find the mass percents. First we let x be the mass of (C2H5)2O and y be the mass of CH3CH2OH. Thus, x + y = 1.005 g or y = 1.005 g – x We then construct a second equation involving x that relates the mass of carbon dioxide formed to the masses of ethanol and diethyl ether., viz.
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1.963 g CO 2
1 mol C2 H 5 2 O 1 mol CO 2 4 mol CO 2 x g (C2 H 5 ) 2 O 44.010 g CO 2 74.123 g C2 H 5 2 O 1 mol C2 H 5 2 O
(1.005 x) g CH3CH 2 OH 0.04460 0.05396 x 0.04363 0.04341x
1 mol CH 3CH 2 OH 2 mol CO 2 46.07 g CH3CH 2 OH 1 mol CH3CH 2 OH x
0.092 g C2 H 5 2 O
0.04460 0.04363 0.092 g C2 H 5 2 O 0.05396 0.04341
100% 9.2% C2 H 5 2 O 1.005 g mixture % CH 3CH 2 OH (by mass) 100.0% – 9.2% (C2 H5 )2 O 90.8% CH 3CH 2 OH % (CH3CH 2 )2 O (by mass)
115. (D) % Cu (by mass)
# g Cu 100 0.7391 g mixture
Let x = the mass, in grams, of CuCl2 in the mixture. Let 0.7391 – x = mass in grams of FeCl3. Total moles AgNO3 = mol AgNO3 react with CuCl2 + mol AgNO3 react with FeCl3 Total moles AgNO3 = 0.8691 L
0.1463 mol = 0.01271 mol AgNO3 1L
To obtain an expression for the amount of AgNO3 consumed by the first reaction, convert from grams of CuCl2 to moles of AgCl: 2 mol AgNO3 1 mol CuCl2 mol AgNO3 that reacts with CuCl2 x g CuCl2 134.45 g CuCl2 1 mol CuCl2 2x mol AgNO3 that reacts with CuCl 2 = 0.014875x 134.45 To obtain an expression for the amount of AgNO3 consumed by the second reaction, convert from grams of FeCl3 to moles of AgNO3: 1 mol FeCl3 3 mol AgNO3 162.21 g FeCl3 1 mol FeCl3 mol AgNO3 that reacts with FeCl3 (0.7391 x) 0.018496 = 0.013668 0.018496x The sum of these two expressions is equal to the total number of moles of AgNO3 : mol AgNO3 that reacts with FeCl3 (0.7391 x) g FeCl3
Total moles AgNO3 = 0.014875x + 0.013668 – 0.018496x = 0.01271 x = 0.2646 g CuCl2 This is the mass of CuCl2 in the mixture. We must now convert this to the mass of Cu in the mixture. 153
Chapter 4: Chemical Reactions
# g Cu 0.2646 g CuCl2
% Cu
1 mol CuCl2 1 mol Cu 63.546 g Cu 0.1253 g Cu 134.45 g CuCl2 1 mol CuCl2 1 mol Cu
0.1253 g Cu 100 % 16.95 % 0.7391 g
116. (D) 1 mol Cu 2 0.766 mol Cu 2 0.307 2.50 mol Cu 2 63.55 g Cu 2 222- 1 mol CrO 4 mol CrO 4 =35.6 g CrO 4 × = 0.307 mol CrO 4 2- ÷0.307 1.00 mol CrO 4 2115.99 g
(a) mol Cu 2 48.7 g Cu 2
1 mol OH mol OH =15.7 g OH × = 0.923 mol OH - ÷0.307 3.01 mol OH 17.01 g OH Empirical formula: Cu 5 (CrO 4 ) 2 (OH)6 -
-
(b) 5 CuSO 4 (aq) 2 K 2 CrO 4 (aq) 6 H 2O (l)
Cu 5 (CrO 4 ) 2 (OH)6 (s) 2 K 2SO 4 (aq) 3 H 2SO 4 (aq) 117. (D) We first need to compute the empirical formula of malonic acid. 1 mol C 34.62 g C 2.883 mol C 2.883 1.000 mol C 12.01 g C 1 mol O 3.88 g H 3.84 mol O 2.883 1.33 mol H 1.01 g H 1 mol O 3.844 mol O 2.883 1.333 mol O 61.50 g O 16.00 g O Multiply each of these mole numbers by 3 to obtain the empirical formula C3H4O4. Combustion reaction: C3 H 4 O 4 (l) 2 O 2 (g) 3 CO 2 (g) 2 H 2 O(l) 118. (D) 2 Al (s) + Fe2O3 → Al2O3 + 2 Fe 1 mol Fe 2 O3 159.69 g Fe 2 O3 1 mol Al mass of Fe 2 O3 2.5 g Al 7.4 g Fe 2O3 needed 26.982 g Al 2 mol Al 2 mol Fe 2 O3 Using 2.5 g Al2O3, only 7.4 g of Fe2O3 needed, but there are 9.5 g available. Therefore, Al is the limiting reagent. 1 mol Al 2 mol Fe 55.85 g Fe 5.2 g Fe 26.982 g Al 2 mol Al 1 mol Fe 2 O3 (b) Mass of excess Fe2O3 = 9.5 g – 7.4 = 2.1 g
(a) Mass of Fe 2.5 g Al
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119. (M) Compute the amount of AgNO3 in the solution on hand and the amount of AgNO3 in the desired solution. the difference is the amount of AgNO3 that must be added; simply convert this amount to a mass. 0.0500 mmol AgNO 3 amount AgNO 3 present 50.00 mL 2.50 mmol AgNO 3 1 mL soln 0.0750 mmol AgNO 3 amount AgNO 3 desired 100.0 mL 7.50 mmol AgNO 3 1 mL soln 1 mol AgNO 3 169.9 g Ag NO 3 mass AgNO 3 (7.50 2.50) mmol AgNO 3 1000 mmol AgNO 3 1 mol AgNO 3
0.850 g AgNO 3 120. (E) The balanced equation for the reaction is: S8(s) + 4 Cl2(g) 4 S2Cl2(l) Both “a” and “b” are consistent with the stoichiometry of this equation. Neither bottom row box is valid. Box (c) does not account for all the S8, since we started out with 3 molecules, but end up with 1 S8 molecule and 4 S2Cl2 molecules. Box (d) shows a yield of 2 S8 molecules and 8 S2Cl2 molecules so we ended up with more sulfur atoms than we started with. This, of course, violates the Law of Conservation of Mass. 121. (D) The pertinent equations are as follows: C3 N 3 OH 3 3 HNCO (g) 8 HNCO + 6 NO 2 7 N 2 + 8 CO 2 + 4 H 2 O
The above mole ratios are used to calculate moles of C3N3(OH)3 assuming 1.00 g of NO2. 1 mol C3 N 3 (OH)3 1 mol NO 2 8 mol HNCO mass C3 N 3 (OH)3 1.00 g NO 2 46.00 g NO 2 6 mol NO 2 3 mol HNCO
129.1 g mol C3 N 3 (OH)3 1.25 g C3 N 3 (OH)3 1 mol C3 N 3 (OH)3
122. (D) The ammonium dichromate reaction is an example of an internal redox reaction. Both the oxidizing agent (Cr2O72-) and the reducing agent (NH4+) are found in the compound in the correct stoichiometry. The third product is N2(g).
2 NH4+ N2 + 8 H+ + 6 e8 H + 6 e + Cr2O72- Cr2O3 + 4 H2O +
-
(NH4)2Cr2O7(s) Cr2O3(s) + 4 H2O(l) + N2(g) 1000 g (NH4)2Cr2O7 ×
1 mol (NH 4 ) 2 Cr2 O 7 1 mol N 2 28.0134 g N 2 = 111.1 g N2 × × 252.065 g (NH 4 ) 2 Cr2 O 7 1 mol (NH 4 ) 2 Cr2 O 7 1 mol N 2
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123. (D) There are many ways one can go about answering this question. We must use all of the most concentrated solution and dilute this solution down using the next most concentrated solution. Hence, start with 345 mL of 01.29 M then add x mL of the 0.775 M solution. The value of x is obtained by solving the following equation. 1.29 M 0.345 L 0.775 M x 1.25 M = (0.345 x) L
1.25 M (0.345 x) L = 1.29 M 0.345 L 0.775 M x 043125 + 1.25x 0.44505 0.775 x Thus, 0.0138 0.475 x x 0.029 L or 29 mL A total of (29 mL + 345 mL) = 374 mL may be prepared this way. 124. (M)
balanced equation: FeTiO3 + 2H2SO4 + 4H2O = TiOSO4 + FeSO4·7H2O 1.00×103 kg FeTiO3 ×
1 kmol FeTiO3 151.725 kg FeTiO3
×
1 kmol FeSO 4 7 H 2 O 1 kmol FeTiO3
×
278.018 kg FeSO 4 7 H 2 O 1 kmol FeSO 4 7 H 2 O
= 1.83×10 kg FeSO 4 7 H 2 O 3
125. (M) mass of Fe 2 O3 = 1.00×103 kg FeSO 4 7 H 2 O ×
1 kmol Fe 2 O3 1 kmol FeSO 4 7 H 2 O 278.018 kg FeSO 4 7 H 2 O 2 kmol FeSO 4 7 H 2 O
159.692 kg Fe 2 O3 = 287 kg kg Fe 2 O3 1 kmol Fe 2 O3
126. (D) (a) 6 CO(NH2)2(l) 6 HNCO(l) + 6 NH3(g) C3N3(NH2)3(l) + 3 CO2(g) (b) mass C3 N 3 (NH 2 ) = 100.0 kg CO(NH 2 ) 2 × ×
127.
1 kmol CO(NH 2 ) 2 60.063 kg CO(NH 2 ) 2
126.121 kg C 3 N 3 (NH 2 )3 1 kmol C3 N 3 (NH 2 )3
×
1 kmol C3 N 3 (NH 2 )3 6 kmol CO(NH 2 ) 2
84 g actual yield 29.4 kg C3 N 3 (NH 2 )3 100 g theoretical yield
(M) (a) 2 C3H6(g) + 2 NH3(g) + 3 O2(g) 2 C3H3N(l) + 6 H2O(l) (b) For every kilogram of propylene we get 0.73 kilogram of acrylonitrile; we can also say that for every gram of propylene we get 0.73 gram of acrylonitrile. One gram of propylene is 0.0238 mol of propylene. The corresponding quantity of NH3 is 0.0238 mol or 0.405 g; then because NH3 and C3H6 are required in the same molar amount (2:2) for the reaction, 0.405 of a kg of NH3 will be required for every 0.73 of a kg of acrylonitrile. To get 1000 kg of acrylonitrile we need, by simple proportion, 1000×(0.405)/0.73 = 555 kg NH3.
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128.
(E) If the sample that was caught is representative of all fish in the lake, there are five marked fish for every 18 fish. Thus, the total number of fish in the lake is determined. 18fish total fish = 100 marked fish = 360 fish 4 102 fish 5 marked fish
FEATURE PROBLEMS 129. (D) (a) The graph obtained is one of two straight lines, meeting at a peak of about 2.50 g Pb(NO3)2, corresponding to about 3.5 g PbI2. Maximum mass of PbI2 (calculated) 1mol KI 1mol PbI 2 461.01g PbI 2 = 2.503g KI = 3.476 g PbI 2 166.0 g KI 2 mol KI 1mol PbI 2 (b) The total quantity of reactant is limited to 5.000 g. If either reactant is in excess, the amount in excess will be “wasted,” because it cannot be used to form product. Thus, we obtain the maximum amount of product when neither reactant is in excess ( i.e., when there is a stoichiometric amount of each present). The balanced chemical equation for this reaction, 2 KI + Pb NO3 2 2 KNO3 + PbI 2 , shows that stoichiometric quantities are two moles of
KI (166.00 g/mol) for each mole of Pb(NO3)2 (331.21 g/mol). If we have 5.000 g total, we can let the mass of KI equal x g, so that the mass of 1mol KI x = Pb NO3 2 5.000 x g. and the amount KI = x g KI × 166.00 g 166.00 amount Pb NO3 2 = 5.000 - x g Pb NO3 2 ×
1mol Pb NO3 2 331.21g
=
5.000 - x 331.21
b g
At the point of stoichiometric balance, amount KI = 2 amount Pb NO 3
2
x 5.000 - x =2× OR 331.21 x = 10.00 × 166.00-332.00 x 166.00 331.21 1660.0 1mol KI x= = 2.503g KI × = 0.01508 mol KI 331.21+332.00 166.00 g KI 5.000 - x = 2.497 g Pb NO3 2 ×
1mol Pb NO3 2
= 0.007539 mol Pb NO3 2 331.21g Pb NO3 2 2.503g KI 1.002 g KI As a mass ratio we have: 2.497 g Pb(NO3 ) 2 1g Pb(NO3 ) 2 0.01508 mol KI 2 mol KI = As a molar ratio we have: 0.007539 mol Pb(NO3 ) 2 1 mol Pb(NO3 ) 2 (c) The molar ratio just determined in part (b) is the same as the ratio of the coefficients for KI and Pb(NO3)2 in the balanced chemical equation. To determine the proportions precisely, we simply use the balanced chemical equation.
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130. (M) The reaction is: 2 NaOH(aq) + Cl2(g) + 2 NH3(aq) N2H4(aq) + 2 NaCl(aq) + 2 H2O(l) 32.045 g per mole reaction 100% = 17.32 % (a) The theoretical maximum = 184.9607 g per mole reaction (b) The actual AE is less owing to side reactions that lower the yield of the product (N2H4). (c) The addition of acetone changes the mechanism, resulting in the elimination of the side reactions between N2H4 and NH2Cl. This result in an increase in yield from ~70% to nearly 100% (d) 2 H2 + N2 N2H4 has an AE of 100%, neglecting any side reactions. 131. (M) The more HCl used, the more impure the sample (compared to NaHCO3, twice as much HCl is needed to neutralize Na2CO3). Sample from trona: 6.93 g sample forms 11.89 g AgCl or 1.72 g AgCl per gram sample. Sample derived from manufactured sodium bicarbonate: 6.78 g sample forms 11.77 g AgCl or 1.74 g AgCl per gram sample. Thus the trona sample is purer (i.e., it has the greater mass percent NaHCO3 ).
SELF-ASSESSMENT EXERCISES 132. (E)
(a) : A reaction proceeds when heated, that is, heat is playing the role of one of the reactants. (b) (aq): The species is fully soluble in water and the reaction has water present to make the dissociation happen (c) stoichiometric coefficient: number of moles of a species that reacts or forms for an ideal balanced equation (d) overall equation: the combination of several related reaction equations to give one reaction 133. (E) (a) Balancing a chemical equations: making the total number of each type of atoms on both sides of the reaction be equal
(b) Making a solution by dilution: making a solution of known concentration by taking a known volume of a more concentrated solution and diluting it to a known volume (c) Limiting reagent: the reactant that is completely consumed in a reaction 134. (E) (a) Chemical formula is the number and kind of each atom constituting a molecule or formula unit, whereas a chemical equation is the relative number of moles of various reactants and the products they yield
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(b) Stoichiometric factor is the mole ratio of reactants reacting with each other while stoichiometric coefficient is the number of moles reactants and products needed to balance a chemical equation (c) Solute is a substance that dissolves in a solvent. A solvent is the component of the solution which determines the solution’s phase (i.e. solid, liquid or gas) (d) Actual yield is the mass of product collected after a reaction, and percent yield is the ratio between actual and theoretical yields multiplied by 100. (e) Consecutive reactions are reactions that occur one after the other in a specific order, simultaneous reactions occur at the same time 135. (E) The answer is (d). Start balancing in the following order: N, O, H and Cu
3 Cu (s) + 8 HNO3 3 Cu(NO3 ) 2 + 4 H 2 O + 2 NO 136. (E) The answer is (d). To determine the number of moles of NH3, used the balanced equation: 2 mol NH 3 # moles NH 3 = 1 mol H 2 O = 0.666 3 mol H 2 O 137. (M) The answer is (a). To determine the number of moles of NH3, use the balanced equation:
2 KMnO4 (s) + 10 KI + 8 H 2SO4 6 K 2SO4 + 2 MnSO4 + 5 I2 + 8 H 2O 5 KMnO 4
6 mol K 2SO 4 = 15 mol K 2SO 4 2 mol KMnO 4
6 mol K 2SO 4 = 3 mol K 2SO 4 10 mol KI 6 mol K 2SO 4 5 H 2SO 4 = 3.75 mol K 2SO 4 8 mol H 2SO 4 5 KI
138. (E) The answer is (a). To determine the answer, used the balanced equation: 2 Ag 2 (CO3 ) (s) 4 Ag + 2 CO 2 + O 2
The ratio between O2 and CO2 is 1:2.
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139. (E) The answer is (c). To solve this, calculate the number of moles of NaNO3.
mol NaNO3 = 1.00 M × 1.00 L = 1.00 mol. 85.0 g 1.00 mol NaNO3 = 85.0 g NaNO3 1 mol Concentration = 85.0 g NaNO3/L. While (b) also technically gives you the correct value at 25 °C, it is not the definition of molarity. 140. (E) The answer is (d). There is no need for calculation, because a starting solution of 0.4 M is needed to make a 0.50 M solution, and the only way to make a more concentrated solution is to evaporate off some of the water. 141. (M) The answer is (b). To determine the molarity, number of moles of LiBr need to be determined first. Therefore, weight% concentration needs to be converted to number of moles with the aid of the density:
Conc. = 5.30% by mass = 5.30 g LiBr/100 g solution Volume of solution = mass / Density = 100 g sol'n
1 mL = 96.15 mL 1.040g
1 mol LiBr = 0.0610 mol 86.84 g LiBr 0.0610 mol 1000 mL Molarity = = 0.635 M 96.15 mL 1L mol LiBr = 5.30 g LiBr
142. (M) The answer is (d). To determine % yield, calculate the theoretical mole yield: 1 mol CCl2 F mol CCl2 F = 2.00 mol CCl 4 = 2.00 mol CCl2 F 1 mol CCl4
% yield =
1.70 mol 100 = 85.0% 2.00 mol
143. (D) To balance the below equations, balance C first, then O and finally H. (a) 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
(b) For this part, we note that 25% of the available carbon atoms in C8H18 form CO and the remainder for CO2. Therefore, 2 C8H18 + 25 O2 → 12 CO2 + 4 CO + 18 H2O
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144. (D) To determine the compound, the number of moles of each compound needs to be determined, which then helps determine number of moles of emitted CO2:
mass CO 2 = 1.000 g CaCO3
1 mol CaCO3 1 mol CO 2 44.0 g CO 2 100.08 g CaCO3 1 mol CaCO3 1 mol CO 2
= 0.4396 g CO 2 mass CO 2 = 1.000 g MgCO3
1 mol MgCO3 1 mol CO 2 44.0 g CO 2 1 mol CO 2 84.30 g MgCO3 1 mol CaCO3
= 0.5219 g CO 2 mass CO 2 = 1.000 g CaCO3 MgCO3
2 mol CO 2 1 mol dolomite 184.38 g dolomite 1 mol dolomite
44.0 g CO 2 = 0.4773 g CO 2 1 mol CO 2
Therefore, dolomite is the compound. 145. (D) The answer is (b). First, the total amount of carbon in our mixture of CH4 and C2H6 must be determined by using the amount of CO2
1 mol CO 2 1 mol C 12.01 g C = 0.758 g C 44.01 g CO2 1 mol CO 2 1 mol C Then, the amounts of CH4 and C2H6 can be determined by making sure that the moles of carbon for both add up to 0.0631: mass of C = 2.776 g CO 2
1 mol CH 4 1 mol C 12.01 g C xg 1 mol CH 4 1 mol C 16.05 g CH 4 1 mol C2 H 6 2 mol C 12.01 g C + (1.000-x) = 0.757 g C (from CO 2 ) 1 mol C 2 H 6 1 mol C 30.08 g C2 H 6 0.748 x + (1-x)(0.798) = 0.757 x = mass of CH 4 = 0.82 g, or 82% of a 1.00 g sample
146. (D) The answer is (c). To do this, perform a stepwise conversion of moles of reactants to moles of products, as shown below:
4.00 mol NH 3
2 mol HNO3 2 mol NO 2 4 mol NO = 2.67 mol HNO3 4 mol NH 3 2 mol NO 3 mol NO 2
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147. (M) To construct a concept map, one must first start with the most general concepts. These concepts contain other more specific concepts discussed in those sections. Looking at sections 4-3, 4-4, and 4-5, it is apparent that the concepts of solution concentration and reaction stoichiometry are the most general ones being discussed. The next stage is to consider more specific concepts that derive from the general ones. In the case of solution concentration, the concept of molarity is a more specific case, and solution dilution yet another. For reaction stoichiometry, the most obvious more specific concept is limiting reagents. Afterwards, link the general and more specific concepts with one or two simple words. For example, molarity is an “expression of” concentration. Take a look at the subsection headings and problems for more refining of the general and specific concepts.
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