QUÍMICA
ESTADOS DE AGREGACIÓN DE LA MA MATERIA TERIA 1. Un gas ideal ocupa un volumen de 500 cm 3 a la presión de 780 mm de Hg. Suponiendo que la masa y la temperatura permanecen constantes y que la presión se reduce a 750 mm de Hg, calcular el volumen final ocupado por el gas.
camente la masa, se verifica la ecuación general P1 V 1 P2 V 2 de los gases ideales: –––––– = ––––––. –––– ––. DespejanDespejanT 1 T 2 P1 V 1 T 2 do, resulta: P2 = –––––––––. En este caso se tiene: V 2 T 1 P1 = 1 atm; T 1 = 40 + 273 = 313 K; V 2 = V 1 + 5 % V 1 = 1,05 V 1; T 2 = 360 + 273 = 633 K.
Sustituyendo los datos numéricos se obtiene:
RESOLUCIÓN: Como es una transformación experimentada por un gas ideal manteniendo constantes la masa y la temperatura, se cumple la ley de Boyle-Mariotte: P1 V 1 = P2 V 2. Despejando, resulta: P1 V 1 V 2 = –––––––. P2
Sustituyendo los valores numéricos del enunciado se obtiene: 780 · 500 750
V 2 = ––––––––– = 520 cm 3.
2. Un matraz cerrado contiene aire a 40 °C y 1 atm. Si se calienta hasta 360 °C y el volumen aumenta en un 5 %, calcular la presión final.
1 · V 1 · 633 = 1,926 atm. P2 = –––––––––––– 1,05 V 1 · 313 3. En una bombona de 12 litros hay oxígeno a 1,4 atm y 310 K. Calcular: a) El número de moles de oxígeno que contiene la bombona. b) La densidad del oxígeno en esas condiciones.
RESOLUCIÓN: a) A partir de la ecuación de estado de d e los gases P V ideales P V = n R T se tiene: n = –––––. Sustitu R T
yendo los valores numéricos del enunciado resulta: 1,4 · 12 oxígeno. 0,082 · 310 b) Para calcular el valor de la densidad del oxígeno en estas condiciones, podemos partir de la n = –––––––––– = 0,661 moles de
RESOLUCIÓN: Como se trata de una transformación experimentada por un gas ideal manteniendo constante úni-
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ecuación de estado de los gases ideales P V = n R T teniendo en cuenta que n = a / M . Sustituyendo, resulta: a a P V = ––– R T o bien: P · M = ––– R T . M V Pero como = a / V V, sustituyendo se obtiene: P M P · M = R T . Despejando: = –––––. R T
Sustituyendo los valores numéricos del enunciado resulta: 1,4 · 32 0,082 · 310
= ––––––––––– = 1,76 g/l
4. Se tiene un gas en condiciones normales. Calcular la presión del gas si la temperatura se duplica permaneciendo constante el volumen.
RESOLUCIÓN: En este caso se trata de una transformación experimentada por un gas ideal manteniendo constantes la masa y el volumen, por tanto, se cumple la P1 P2 ley de Charles y Gay-Lussac: ––– = –––. DespeP T T 1
T 1
T 2
1 2 jando, resulta: P2 = ––––––. Sustituyendo los va-
lores del enunciado, se tiene: 1 · 2T 1 = 2 atm. P2 = ––––––– T 1
5. Un recipiente contiene 150 cm 3 de un gas ideal a 42 °C y 714 mm de Hg. Si después de una serie de transformaciones el volumen del gas es de 102 cm3 a la presión de 830 mm de Hg, calcular la temperatura final del gas, expresada en grados
P1 V 1 P2 V 2 ––––––– = –––––––. T 1
Despejando, resulta:
T 2
P2 V 2 T 1 T 2 = ––––––––. P1 V 1
En este caso se tiene: T 1 = 42 + 273 = 315 K. Sustituyendo los valores numéricos resulta: 830 · 102 · 315 714 · 150 = 249 K, es decir, 249 – 273 = –24 °C. T 2 = –––––––––––––– =
6. El cuerpo humano produce en la respiración una media de 40 g de dióxido de carbono cada hora. Calcular el volumen de dióxido de carbono medido a 27 °C y 101 kPa que se produce en la respiración de una persona durante todo un día. Tómese R = 8,31 J · K –1 mol–1.
RESOLUCIÓN: A partir de la ecuación de estado de los gases ideales P V = n R T , teniendo en cuenta que n = a / M resulta: a P V = –––– R T . M
Despejando, se obtiene: a R T V = ––––––. P M
Sustituyendo los valores numéricos: 40 · 8,31 · 300 m3 V = –––––––––––––– = 0,02244 ––––– 44 · 101.000 hora Es decir, 3 m 1.000 l · ––––––––– 24 horas = 0,0224 –––––– · ––––––– hora 1 m3 1 día litros = 538,5 –––––– día
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RESOLUCIÓN: Empleando los factores de conversión correspondientes resulta: 1 mol Xe · ––––––––– 22,9 KJ = 52,3 KJ. 300 g Xe · ––––––––––– 131,29 g Xe 1 mol Xe
RESOLUCIÓN: A partir de la ecuación de estado de los gases ideales P V = n R T , teniendo en cuenta que n = a / M resulta: a aRT P V = ––– R T . Despejando, se obtiene: M = ––––– M P V
Sustituyendo los valores numéricos, resulta: 8. Uno de los compuestos formado por flúor, cloro y carbono que destruye la capa de ozono de la atmósfera es el CClF 3. Este compuesto tiene un punto de ebullición de 23,8 °C, su calor de vaporización es de 180,2 J/g y su densidad en estado líquido es de 1,47 g/cm 3. Calcular: a) El volumen ocupado por 10 g del compuesto medidos a 100 °C y 100 kPa. b) La energía necesaria para vaporizar totalmente 2 moles del compuesto sin variar la temperatura. Tómese R = 8,31 J · K –1 mol–1.
RESOLUCIÓN: a) A partir de la ecuación de estado de los gases ideales P V = n R T , teniendo en cuenta que n = a / M a R T . Despejando, se obtiene: resulta: P V = –––– M
a R T V = ––––––. Sustituyendo los valores numéricos, P M
resulta:
0,48 · 8,31 · 300 100.000 · 400 · 10
M = ––––––––––––––––– –6 = 29,9 g/mol
10. A temperatura y presión ordinarias el propano es un gas. Calcular: a) Su densidad en condiciones normales. b) Su densidad a 100 °C y 10 kPa.
RESOLUCIÓN: a) Se entiende por condiciones normales P = 1 atm y T = 0 °C = 273 K. A partir de la ecuación de estado de los gases ideales P V = n R T , teniendo en cuenta que n = a / M resulta: a P V = ––– R T . M V , sustituyendo = a / V
Pero como: P M = R T . Despejando:
se tiene:
P M = ––––– R T
Sustituyendo los valores numéricos resulta: 10 · 8,31 · 373 100.000 · 104,5
V = ––––––––––––– =
103 litros = 2,97 litros. = 2,97 · 10–3 m3 · –––––––––– 1 m3 b) Se tiene: 104,5 g CClF3 · ––––––––– 180,2 J =. 2 moles · ––––––––––––– 1 mol 1 g CClF3
10.000 · 44 8,31 · 373 b) Procediendo análogamente, se tiene: P M 10.000 · 44 = ––––– = ––––––––––– = 8,31 · 373 R T = –––––––––– = 1,96 g/l.
3 g g 1 m = 141,9 –––3 · ––––––– Ӎ 0,14 –––. m 1.000 l l
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RESOLUCIÓN: En este caso se trata de una transformación experimentada por un gas ideal manteniendo constantes la masa y la presión, presión , por tanto, se cumple la
Sustituyendo los valores numéricos, resulta: 141.284 · 1 101.300
V 2 = –––––––––– = 1,4 cm 3.
V 1 V 2 ley de Charles y Gay-Lussac: ––– = –––. DespeT 1
T 2
T 1 V 2 jando,, resulta jando resulta:: T 2 = –––––– Sustituyendo los valoV 1
res del enunciado, se tiene: 288 · 100 T 2 = ––––––––– = 1.152 K, 25 es decir: 1.152 – 273 = 879 °C. 12. Una burbuja de aire situada a 4 m de profundidad bajo el agua del mar tiene un volumen de 1 cm3. Sabiendo que la densidad del agua del mar es de 1,02 g/cm 3 y suponiendo que la temperatura permanece constante, calcular el volumen de la burbuja en la superficie. Tómese para la presión atmosférica el valor de 101.300 Pa.
13. Calcular el volumen ocupado por 100 g de nitrógeno en condiciones normales.
RESOLUCIÓN: En condiciones normales de presión y temperatura 1 mol de un gas ideal ocupa un volumen de 22,4 litros. Así pues, en este caso, se tiene: 1 mol N2 · –––––––––––– 22,4 l 100 g N2 = –––––––– = 80 litros. 28 g N2 1 mol N2 (c.n.) 14. Un manómetro indica 42 lb/pulg 2 en un tanque de aire comprimido. Hallar su presión absoluta en mmHg y kPa.
RESOLUCIÓN: RESOLUCIÓN: Como se trata de una transformación experimentada por un gas ideal manteniendo constantes la masa y la temperatura, se s e cumple la ley de Boyle-Mariotte: P1 V 1 = P2 V 2. Si despejamos, resulta: P1 V 1 Ahora bien, en este caso, la presión V 2 = ––––––. P2
inicial será la suma de la presión atmosférica más la presión hidrostática que soporta la burbuja. Es decir: P1 = Patm + Ph. La presión hidrostática viene dada por la expresión: Ph = = g h. Sustituyendo los valores numéricos se obtiene: g m · 4 m · ––––––– 1 kg · Ph = 1,02 ––––3 · 9,8 ––– cm s2 1.000 g
P. absoluta = P. manométrica + P. barométrica Pab = 42–––– lb + 14,7 –––– lb pulg2 pulg2 Pab = 56,7 –––– lb pulg2 1at = 14, 14,77–––– lb = 760 mmH mmHgg = 1,0 1,013 13 . 105 kPa pulg2 luego 56,7 –––– lb pulg2 Pa = ––––––––– 14,7 lb
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15. Una masa gaseosa ocupa un volumen de 0,2 lts a 212 °F. Hallar el volumen en pies 3 que ocupará a 0 °C en un proceso isobárico.
RESOLUCIÓN: Por Charles ––– V1 = ––– V2 T1 T2 Datos: V1 = 0,2 L T1 = 212°F · 460 = 672°R (temp. abst.) T2 = 0°C = 32°F + 460 = 492°R T2 De la fórmula V2 = V1 . ––– T1 Reemplazando valores 492°R V2 = 0,2L . –––––– 672°R V2 = 0,146 L Pero 1 pie3 < > 28,32 lt 0,146 L pie3 luego V2 = –––––– 28,32 L V2 = 0,0052 pie3 16. Cierta masa gaseosa ocupa un volumen de 500 L a 35 °C y 770 torr (770 mmHg); se comprime dentro de un tanque de 100 L de capacidad a la presión de 5 at. Hallar la temperatura final del gas.
RESOLUCIÓN:
P1 = 770 torr P2 = 5at . 760 torr = 3 800 ––– at Se sabe : P 1 V1 P 2 V2 ––––– = ––––– T1 T2 770 3 800 · 100 · 500 = –––––––––– –––––––– 308 T2 3 800 · 100 · 308 T2 = –––––––––––––– 770 · 500 T2 = 304°K T2 = 304 - 273 T2 = 31°C 17. Se tienen 5 lts de anhídrido sulfuroso SO 2 a 18 °C y 1 600 mmHg. Calcular su volumen en condiciones normales (CN).
RESOLUCIÓN: Datos: V1 = 5 L T1 = 18 + 273 = 291°K P1 = 1 600 mmHg
Vn = ? Tn = 273°K Pn = 760 mmHg
Por la ley general de los gases P1V 1 PnV n –––– = ––––– T 1 T n
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18. Calcular la masa de 150 mL de HCI gaseoso que se encuentra en condiciones normales (CN).
RESOLUCIÓN: Se sabe que 1 mol de gas ocupa siempre 22.4 lts, también 1 mol de HCL (1+35.5) = 36,5 g
20. 2,04 g de una sustancia pura se vaporiza a 55°C y 780 torr de presión. El volumen obtenido en estas condiciones es de 230 cm3. Hallar el peso molecular de dicha sustancia.
RESOLUCIÓN: Se sabe PV = RTR PV = RT W/M
luego:
RTW PV
M = –––––
En 22.4 lts de HCI –––––– hay 36,5g x En 0.15 lts de HCI –––––– x =
0,15 . 36,5 ––––––––– 22,4
x = 0,244 g
19. Calcular el peso molecular de un gas si 455 cm3 del mismo en condiciones normales pesan 2,43 . 10-2 N.
RESOLUCIÓN: Se sabe m=
P = V = W= T = R =
Peso = m . g
780 torr (mmHg) 0,230 L 2,04 2, 04 g 55°C + 273 = 328°K 62,4 ––––––––– mmHg x L °K x mol
Reemplazando M =
62,4 . 328 . 2,04 –––––––––––––– 780 . 0,230
M = 232,73 g/mol
Peso –––– g
-2 N 2,43 . 10 m = ––––––––––– 9,8 m/s2
m = 2,48 . 10-3 kg m = 2,48 g
luego
De acuerdo a los datos:
¿Cuánt ántos os átom átomos os de de P y O hay hay en 0,2 . 10-3 21. ¿Cu m3 de P205 a 37 °C y 106,6 . 103 Pa. RESOLUCIÓN: Datos: P = 106,6 . 103 Pa V = 0,2 . 10-3m3
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PV = n ––– RT 106,6 . 103 . 0,2 x 10-3 = n ––––––––––––––––––– 8,31 · 310 8,276 . 10-3 moles = n Calculamos el # de moléculas 8 276 . 10-3 . 6 023 . 1023 # moléculas de P 205 = 4,98 . 10 21 # átomos de P = 2 . 4,98 . 10 21 = 9,96 . 1021 # átomos de O = 5 . 4,98 . 10 21 = 24,90 . 1021 22. ¿Qué relación existe entre las velocidades de difusión del H2 y CO2 que pasan a través de un fino agujero?
RESOLUCIÓN: Por la ley de Graham V1 ͙ M2 (M = Masa molecular) ෆ –– = ––––– V2 ͙ M1 (V = Velocidad de difusión) ෆ Sea V1 = V (H2) V2 = V (CO2)
Volumen Nota: Velocida elocidadd de difusió difusiónn = –––––––– Tiempo RESOLUCIÓN: Se sabe V1 ͙ M2 ෆ –– = ––––– V2 ͙ M1 ෆ Datos: V(CO2) V(O2) M(CO2) M(O2)
= = = =
100 cm3 /h ? 44 16 . 2 = 32
V(CO2) ͙ M(O2) –––––– = ––––––––– V(O2) ͙ ෆළ M(CO2) ෆළ
100 = ––––– ෆ ͙ 32 ––––– V(O2) ͙ ෆ 44 V(O2) = 117,2 cm 3 /h Pero por ser en 5’ que equivale a 1/12 de hora 117,2 Resp. = ––––– 12 V2 = 9,77 cm3
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la masa final de la disolución será: 80 + 6 = = 86 gramos. Por consiguiente, el porcentaje en peso de cloruro potásico de la disolución resultante será: 15,6 · 100 = 18,1 % ––––– 86
2. En una disolución de ácido sulfúrico al 26 % en peso y densidad 1,19 g/cm 3, calcular: a) La normalidad de la disolución. b) La molaridad de la disolución. c) El volumen de agua que habrá que añadir a 0,1 litros de la disolución anterior para obtener una disolución 3 N de dicho ácido. RESOLUCIÓN: a) Como m = V · , la masa de 1 litro de esta disolución será de: 1.000 cm3 · 1,19 g/cm3 = 1.190 gramos. Como su concentración es del 26 %, la cantidad de ácido sulfúrico que contiene es: 1.190 · 26 % = 309,4 gramos de ácido sulfúrico. Como esta cantidad de ácido sulfúrico está disuelta en 1 litro de disolución, pasada a equivalentes y a moles será, por definición, la normalidad y la molaridad de la disolución. En el caso del ácido sulfúrico, que posee 2 hidrógenos, 1 equivalente = 1/2 mol = 98/2 = 49 gramos. Por tanto, resulta:
1 eq
Despejando resulta: 1.000 · 0,632 x = –––––––––––– = 211 cm 3 3 de disolución. Así, se deberán añadir 211 – 100 = = 111 cm3 de agua a los 100 cm 3 de disolución 6,32 N. 3. Se dispone de una disolución 0,1 M de KCl a partir de la cual se desea preparar una disolución 0,002 M de esta sal. Calcular los ml de la primera disolución que se precisan para preparar 250 ml de la segunda. s egunda.
RESOLUCIÓN: Los 250 ml (0,25 litros) de disolución 0,002 M que queremos preparar, contendrán 0,25 · 0,002 0, 002 = = 5 · 10–4 moles de KCl. El volumen x de disolución más concentrada, 0,1 M, que contiene estos mismos moles es posible calcularlo a partir de –4 mol 0 , 1 m o l K C l 5 · 1 0 ––––––––––––––––– = ––––––––––––. 1.000 ml disolución x
Despejando, resulta: 1.000 · 5 · 10 –4 0,1
x = –––––––––––––– = 5 ml de disolución 0,1 M.
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DATO: La presión del vapor del agua a 25 °C es de 23,8 mm Hg. RESOLUCIÓN: a) La masa de 1 litro de disolución de ácido sulfúrico comercial será: m = V . Sustituyendo los valores numéricos resulta: m = 1.000 cm3 · 1,83 g/cm3 = = 1.830 gramos. De estos gramos, el 91%, es decir 1.830 · 91% = 1.665 gramos, son de ácido sulfúrico. Teniendo Teniendo en cuenta que la masa molar del ácido sulfúrico es de 98 g/mol, los 1.665 gramos de 1 mol = ácido sulfúrico equivaldrán a: 1.665 g · –––––– 98 g = 17 moles de ácido sulfúrico. Como los 17 moles de ácido sulfúrico están disueltos en 1 litro de disolución, ésta será, por definición, 17 M. b) 1 litro de disolución 0,5 N contiene 0,5 equivalentes de ácido sulfúrico. En el caso del ácido sulfúrico, que posee 2 hidrógenos, 1 equivalente = = 1/2 mol = 98/2 = 49 gramos. Por tanto, 0,5 equivalentes serán 0,25 moles. Como queremos preparar 1 litro de disolución 0,5 N, calcularemos el volumen x de disolución comercial (17 M) que contiene 0,25 moles, que será el volumen que deberemos tomar tomar.. Se tiene: 17 mol H2SO4 = –––––––––. 0,25 mol –––––––––––––––– x 1.000 cm3 disoluc. Despejando resulta: 1.000 · 0,25 x = –––––––––– = 14,7 cm 3 de disolución 17 M. 17
Sustituyendo valores numéricos: 721,2 · 390 · 273 V 0 = –––––––––––––––. 760 · (273 + 25) Es decir: V 0 = 339 cm3 de hidrógeno medidos en condiciones normales. En consecuencia, se verifica: 22.400 cm3 (c.n.) = ––––––––. 339 c m 3 ––––––––––––––– x 1 mol Despejando, resulta: x = 0,015 moles de hidrógeno. Como la molécula de hidrógeno contiene 2 átomos de hidrógeno, 1 mol de hidrógeno contiene 2 equivalentes. Entonces, los 0,015 moles de hidrógeno serán 0,03 equivalentes. Teniendo en cuenta que, por definición, un número de equivalentes determinado de un metal desplaza de un ácido el mismo número de equivalentes de hidrógeno, habrá también 0,03 equivalentes de metal. Como se utilizó 1 gramo de metal, 1 gramo coincidirá con 0,03 equivalentes y, por tanto, el peso equivalente del metal se podrá calcular a partir de la expresión: 0,03 eq = ––––– 1 eq ––––––– x 1g Despejando, resulta: x = 33,3 gramos. Por consiguiente, el peso equivalente buscado es 33,3 gramos. 5. En el análisis de una blenda, en la que todo el azufre se encuentra combinado como ZnS, se
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Como todo el azufre contenido en el ZnS aparece en el BaSO4, teniendo en cuenta los pesos atómicos se verifica verifica:: 233,3 g BaSO4 (1mol) = –––––––. 1,878 g –––––––––––––––––––– 97,4 g ZnS (1 mol) x Despejando: 97,4 · 1,878 x = ––––––––––– = 0,7840 g de ZnS. 233,3 Por tanto, habrá 0,7840 gramos de ZnS en 0,9364 gramos de blenda, por lo que la riqueza de esta blenda será: 0,7840 · 100 = 83,7 %. –––––––– 0,9364
6. Calcular el volumen de hidrógeno medido a 50 °C y 1,2 atm de presión que se obtiene al añadir 75 ml de HCl 0,5 M a 10 g de Al.
RESOLUCIÓN: La ecuación correspondiente al proceso que tiene lugar es: 2 Al + 6 HCl
→
2 AlCl3 + 3 H2.
En primer lugar calcularemos el número de
3 · 22,4 l (c.n.) H 2 (3 mol) = 0,0375 mol · –––––––––––––––––––––– 6 mol = 0,42 l H2 (c.n.). De acuerdo con la ecuación general de los gases perfectos se tiene: P0 V 0 P V –––––– = –––––. T 0
T
Despejando resulta: P0 V 0 T V = ––––––––. P T 0
Sustituyendo los valores numéricos se obtiene: 1 · 0,42 · 323 1,2 · 273
V = –––––––––––– = 0,41 l H 2.
7. Calcular los litros de hidrógeno medidos a 750 mm de Hg de presión y 30 °C de temperatura que se pueden obtener atacando 75 g de Zn metálico del 90 % de riqueza con ácido sulfúrico.
RESOLUCIÓN: Teniendo en cuenta la riqueza de la muestra, la cantidad de Zn puro de que se dispone es: 75 · 90 % = 67,5 gramos. La ecuación correspondiente al proceso es: Zn + H2SO4 → ZnSO4 + H2. De acuerdo con la relación estequiométrica del
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8. Al añadir agua al carburo de calcio se obtiene hidróxido de calcio y acetileno. Calcular los gramos de agua necesarios para obtener 2 litros de acetileno a 27 °C y 1 atm.
1 mol = 0,5988 Moles de N: 8,383 g · –––––– 14 g
RESOLUCIÓN: La ecuación química correspondiente al proceso es: CaC2 + 2 H2O → Ca(OH)2 + C2H2.
Dividiendo los resultados anteriores por 0,5988 se obtiene la relación
Aplicando ahora la ecuación general de los gases, PV P0 V 0 ––––– = –––––– se puede calcular el volumen de T T 0
acetileno en condiciones normales, sustituyendo 760 · 2 = –––––––. 760 V 0 los valores numéricos: ––––––––––– (273 + 27) 273 Despejando se obtiene: 760 · 2 · 273 V 0 = –––––––––––– = 1,82 l C 2H2 (c.n.) 760 · 300 La cantidad de agua necesaria para producir este volumen de acetileno, de acuerdo con la estequiometría del proceso será: 2 · 18 g H2O (2 moles) = 1,82 l C2H2 (c.n.) · –––––––––––––––––––––– 22,4 l C2H2 (c.n.) (1 mol) = 2,9 g H2O
1 mol = 0,5988 Moles de O: 9,581 g · –––––– 16 g
6,2875 = 10,5 Moles de C: ––––––– 0,5988 6,587 = 11 Moles de H: ––––––– 0,5988 0,5988 = 1 Moles de N: ––––––– 0,5988 0,5988 = 1 Moles de O: ––––––– 0,5988 Multiplicando los resultados anteriores por 2, para que todos los números que aparecen en la relación sean enteros se obtiene: Moles de C: 10,5 · 2 = 21 Moles de H: 11 · 2 = 22 Moles de N: 1 · 2 = 2 Moles de O: 1 · 2 = 2 Por tanto, la fórmula empírica será:
9. La estricnina es un poderoso veneno utilizado como raticida. La composición porcentual de
(C21H22O2N2)n.
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2 NaNO3 + H2SO4 → Na2SO4 + 2 HNO3. Por la estequiometría del proceso se tiene: 98 g H2SO4 (1 mol) 6,3 kg HNO3 · ––––––––––––––––––––––––– = 70 2 · 63 g HNO 3 · –––– (2 mol) 100 = 7,0 kg H2SO4 = 7.000 g H2SO4.
Dividiendo los resultados anteriores por 0,0619 se obtiene la relación Moles de C: 0,0619 : 0,0619 = 1 Moles de H: 0,0622 : 0,0619 Ӎ 1 Moles de Cl: 0,0619 : 0,0619 = 1 Por consiguiente, la fórmula empírica del lindano es (CHCl)n.
Así pues, el volumen de la disolución comercial que contiene los 7,0 kg de ácido sulfúrico que se necesita para efectuar el proceso es:
12. ¿Cuántos gramos de solución al 15% de NaCl se necesitan para extraer 38g de NaCl?
1 l disoluc. = 7.000 g H2SO4 · ––––––––––––– 1,350 g H2SO4 = 5.180 l disolución H2SO4.
RESOLUCIÓN: Se sabe: % en peso = –––––––––––––– Peso soluto . 100 Peso de solución Reemplazando valores:
11. El insecticida lindano contiene C, H y Cl. Al quemar una muestra de 3,00 g del mismo se obtienen 2,72 g de dióxido de carbono y 0,56 g de agua. Determinar la fórmula empírica de este compuesto.
RESOLUCIÓN:
38 . 100 15 = ––– x x =
38 . 100 ––––––––
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Luego En 663g de Al2(SO4)3.18 H2O existen existen 54 g de Al+3
x
x
––––– 2 g de Al+3
De acuerdo a la ecuación
2 . 663 = –––––– 54
x = 24,55 g de
BaCI2 + H2SO4 → BaSO4 + 2HCl Masa M 208,4 98 233,4 73
Al2(SO4)3 . 18 H2O
98 g de H2SO4 producen 233,4 g de BaSO BaSO4 x g de BaSO4 ––––––– 4,08 g de BaSO4 luego
14. Una solución concentrada de H 2SO4 tiene 95% en peso de H2SO4 puro y una densidad de 1,84 g/cm3. ¿Qué volumen de esta solución contiene 360g de H2SO4 puro?
RESOLUCIÓN: (Nota: % en peso = % en masa) Partiendo del concepto de densidad:
98 . 4,08 233,4
x = –––––––– x = 1,71 g de
H2SO4
Cálculo del % de pureza 1,71 . 100 = 42,75% –––– 4
Si δ = 1,84 g/cm3 significa que: En 1 cm 3 hay 1,84 g de masa pero
16. Se prepara Cl2 por reacción de MnO2 con H SO y NaCl. ¿Qué masa de MnO y NaCl y
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Cálculo de la masa de NaCl: 2 moles de NaCl (117g) –––– 1 mol Cl 2 a C.N. 117 g producen 22,4 L x ––––––– 4,48 L 117 . 4,48 22,4
x = –––––––––– x
= 23,4 g de NaCl
Cálculo del volumen de solución de H2SO4 18 M 3 mol de H2SO4 ––––––––– 22,4 L a C.N. x ––––––––– 4,48 L 4,48 . 3 x = ––––––– 22,4 x = 0,6 moles de
H2SO4 puro
De acuerdo al enunciado el H SO es 18 M, por
Eq = 162,4 ––––– 3 Eq = 54,1 Masa # Eq = –––––––––– Eq. químico 5,45 # Eq = ––––– 5,41 # Eq = 0,10073 0,10073 Por tanto N = –––––––– 0,060 N = 1,68
18. Calcular la masa de soda caústica que se necesita para preparar 5L de una disolución al 20% y de densidad 1 219 g/cm 3. Calcule además
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M (M = masa molecular) Eq = ––––––– # de OH 40 Eq = –––– 1 Eq = 40 entonces # de Eq = ––––––––– masa total Eq 1 219 # de Eq = ––––– 40 # Eq = 30 475 en 5L
recordar que para que una neutralización se lleve a cabo, los # Eq de ambas soluciones deberán ser iguales. # EqA = # EqB pudiendo presentarse la expresión anterior de las siguientes maneras: MA = –––– MB ………………… (1) –––– EqA EqB NAVA = NBVB ………………… (2) RESOLUCIÓN: En nuestro caso
# Eq . soluto N = ––––––––––– # L solución
#EqH2SO4 = #EqNaOH
30 475 N = ––––––– 5
mH2SO4 = ––––––––– mNaOH …………… (3) –––––––– Eq H2SO4 Eq NaOH
N 6 095 Eq/L
Masa molecular
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21. En una mezcla de gases a 20 °C, las presiones ejercidas por sus componentes son: H 2 = 200 mmHg; CO2 = 150 mmHg CH4 = 320 mmHg y C2H4 = 105 mmHg. Calcule la presión total de la mezcla y el porcentaje de H 2 en volumen.
Recordando las leyes de Dalton respecto a la mezcla de gases: *Presión total = Suma de las presiones parciales de los componentes. *Porcentaje molar = Porcentaje volumétrico. *Las presiones parciales son proporcionales a sus respectivos % molares. *Las presiones parciales son proporcionales a sus porcentajes volumétricos. RESOLUCIÓN: En nuestro caso: a) PT
P
= PpH2 + PpCO2 + PpCH4 + PpC2H4 200 mmHg + 150 mmHg + 320 mmHg +105 mmHg
torr; 80 cm3 de O2 a la presión de 400 torr y 700 cm3 de N2 a la presión de 200 torr. Calcular la presión total de la mezcla. De acuerdo al enunciado, calcularemos las presiones finales de manera independiente; por ser un proceso isotérmico, aplicaremos la ley de Boyle para cada caso. RESOLUCIÓN: Se sabe P1V1 = P2V2 1V1 = P de donde: P–––– 2 V2 Para el H2
P1 = 700 torr V1 = 20 2000 cm cm3 V2 = 25 2500 cm cm3 . 200 cm3 700 torr P2 = ––––––––––––––– 250 cm3 P2 = 560 torr Para el O
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23. A 25 °C y a 102 kPa se indica 1,5 m3 de O2, húmedo al 90%. ¿Qué volumen ocupará si se procede a un secado parcial hasta 30% de H.R. a una temperatura de 20°C y presión 80 kPa?
RESOLUCIÓN: Se sabe que: Ppva a 25°C = 23,80 mmHg y Ppva a 20°C = 17,54 mmHg Nota: La ley general de los gases, como todas las leyes de los mismos, es aplicable a gases ideales y no a gases húmedos, que no son ideales; para que se cumpla deberá trabajarse con gases secos por lo que deberá restarse a la presión total del
T1 V2 P2 T2
= = = =
25° + 27 273 = 298°K ? 80 – 0,70 = 79, 79,3 kPa 20°C 20 °C + 27 2733 = 29 2933 °K . 1,5 . V2 99,15 79,3 ––––––––– = ––––––– 298 293 99,15 . 1,5 . 293 = V –––––––––––––– 2 298 . 79,3 1,84 m3 = V2
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ESTRUCTURA ATÓMICA 1. Un electrón excitado de un átomo de hidrógeno vuelve al estado fundamental y emite una radiación electromagnética de 180 nm. Calcular: a) La frecuencia de la radiación. b) La diferencia de energía entre los dos niveles energéticos expresada en joules. Tómese: h = 6,62 · 10–34 J · s; c = 3 · 10 8 m/s.
RESOLUCIÓN: a) Se trata de un problema sobre el espectro de emisión del átomo de hidrógeno. Mediante la ecuación que relaciona la frecuencia y la longitud de onda encontraremos la frecuencia. Es decir: =
tener en cuenta que la energía de la radiación emitida equivale a la diferencia de energía entre los dos estados entre los que se produce la transición. La energía de la radiación emitida, de acuerdo con la ecuación de Planck, es: E = h y equivale a la diferencia de energía entre los dos estados entre los que se produce la transición, ya que un átomo emite radiación electromagnética cuando un electrón pasa de un nivel de más energía a otro de menos energía y la absorbe en la transición inversa. Teniendo en cuenta que = c / se obtiene: E =
hc ––––. Sustituyendo los valores numéricos,
resulta: –34 · 3 · 10 8 6,6 · 10 E = –––––––––––––––– = 3,36 · 10 –19 J.
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4. Un rayo gamma tiene una longitud de onda de 0,01 Å. Calcular la energía correspondiente a sus fotones.
RESOLUCIÓN: La energía de los fotones viene dada por la ecua-
h = –––––. mv
Sustituyendo los valores numéricos se obtiene: –34 6,63 · 10 = 6,63 · 10 –9 m. = ––––––––––– –31 6 10 · 10
hc ción de Planck: E = h = ––––. Sustituyendo los
valores numéricos resulta: 6,63 · 10–34 · 3 · 108 0,01 · 10
= 1,989 · 10 –13 J. E = ––––––––––––––––– –10
5. Calcular la masa equivalente de un fo-
7. Una estación de radio emite con una frecuencia de 1.000 kc · s –1. Calcular la energía que corresponde a los cuantos de esta radiación electromagnética.
RESOLUCIÓN: En primer lugar, convertiremos el valor de la fre
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tante de Rydberg vale 109.677,581 cm–1, calcular el valor de m. RESOLUCIÓN: De la ecuación de Rydberg se tiene:
(
1
1
Por la ecuación de Planck, se tiene: E = h . Ahor Ah oraa bi bien en:: = c ¯. Combinando ambas expresiones resulta: E = h c ¯. Sustituyendo los valores numéricos se obtiene: 7 144
E = 6,63 · 10 –34 · 3 · 10 8 · 10.967.758,1 · –––– =
)
– –––2 . ¯ = R ––– n2 m
= 1,06 · 10–19 J Es decir:
Como la línea pertenece a la serie de Lyman, se tiene n = 1. Sustituyendo en la ecuación de Rydberg resulta:
1 eV = 0,662 eV. 1,06 · 10–19 J · –––––––––– 1,6 · 10–19 J
1 1 ¯ ¯ = R 1 – –––2 . De donde: ––– = 1 – –––. m R m2
11. Calcular la longitud de onda asociada a un neutrón que se mueve a una velocidad de
(
)
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se trata de n fotones, la ecuación anterior se transforma en E = n h . Pero como = c / , combinando ambas ecuaciones resulta: nhc E = ––––––.
Despejando: 1 1 = –––––––– = –––––––––––––––– = R · 0,21 10967758,1 · 0,21 = 4,34 · 10–7 m. La energía de la radiación emitida vendrá dada hc por la ecuación de Planck: E = h = ––––. Pero
Despejando:
E n = –––––. hc
Sustituyendo los valores numéricos resulta: 1 · 580 · 10 –9 n = –––––––––––––––––– = 6,626 · 10–34 · 3 · 108 = 2,9 · 1018 fotones por segundo
esta energía ha sido suministrada por la energía cinética del electrón al chocar con el átomo de hihc drógeno. Por tanto, debe cumplirse que: E c = ––––.
Sustituyendo los valores numéricos resulta: –34 · 3 · 10 8 6,626 · 10 E c = –––––––––––––––––– = 4,578 · 10 –19 J, –7
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15. La energía de la masa en reposo de un cuerpo en reposo es de 1,35 . 1015 joules. ¿A qué velocidad debe moverse dicho cuerpo para que su masa se incremente en 3g?
RESOLUCIÓN: Calculamos la masa en reposo
16. Reducir 10 cm/s2 a pulg/min2.
RESOLUCIÓN: 10 cm pulg ––2 = x –––– s min2
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RESOLUCIÓN: Se sabe: C = f C f
= –––
Masas atómicas: 1H = 1,0078 g; 4He = 2,0026 g; 1 2 RESOLUCIÓN: De la reacción tenemos: mB + mH = 3mHe + mE
11B 5
= 11,011 g
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Por la Ecuación de Rydberg:
Pat = Pa1%1 + Pa2%2 + Pa3%3
1 = 109678 cm -1 –– 1 - –– 1 ––– 4 9
. . . Pat = 28 92,27 –––– + 29 –4,68 ––– + 30 –3,05 ––– 100 100 100
(
Luego:
)
Pat = 28,107
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RESOLUCIÓN: A partir de los números atómicos se pueden determinar las configuraciones electrónicas de los elementos del enunciado: Ca 1s2
S 1s2
Mg 1s2
O 1s2
3s2 3 p4
3s2 3 p6 4s2
De las configuraciones electrónicas se deduce que el elemento de número atómico Z = 16 se encuentra en el grupo VIa (elementos anfígenos), en el tercer período. Se trata del azufre. Análogamente,
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ro atómico. Para un período del Sistema Periódico, la afinidad electrónica aumenta hacia la derecha, excluyendo los gases nobles. Es decir, es mínima en el metal alcalino y máxima en el halógeno correspondiente. Con este criterio, resulta inmediato observar que el K, que es un metal alcalino, tendrá una afinidad electrónica
5. La afinidad electrónica del yodo es –3,06 eV/átomo. eV/á tomo. Calcular la energía liberada al ionizar 63,45 g de átomos de yodo gaseoso cuyos átomos se encuentran en el estado fundamental.
RESOLUCIÓN:
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EqFe+3
= 55,85 ––––– …… luego EqFe+3 = 18,62 3
Como en cada caso se depositan 0,2 equivalentes, se tendrá entonces: Ag : 0,2 . 107,9 ……… 21,58 g de Ag Zn : 0,2 . 32,69 ……… 6,538 g de Zn Fe : 0,2 . 18,62 ……… 3,724 g de Fe
REACCIONES DE PRECIPI PRECIPIT TACIÓN Y FORMACIÓN DE COMPLEJOS 1. Se tiene una disolución de nitrato de plata 0,001 M. Sabiendo que la constante del producto de solubilidad del cloruro de plata es 2 · 10 –10, calcular la concentración mínima de iones cloruro para que se forme precipitado de cloruro de plata.
RESOLUCIÓN: El proceso que tiene lugar es el siguiente: AgCl(s) Cl–(aq) + Ag+(aq).
b) Indicar si la solubilidad
del yodato de bario será mayor, menor o igual en una solución 0,3 M de cloruro de bario. RESOLUCIÓN: a) En 250 ml de disolución acuosa saturada de yodato de bario hay disueltos 87,85 mg de dicha sal. Por tanto, la concentración de la disolución será: –3 g Ba(IO ) 87,85 · 10 1 mol Ba(IO3)2 = 3 2 · ––––––––––––––– –––––––––––––––––––– 0,25 l 487,3 g Ba(IO3)2 = 7,2 · 10–4 M.
El equilibrio de solubilidad que tiene lugar es el siguiente: Ba(IO3)2 (s) Inicial Equilibrio
s
0
Ba2+(aq) + 2 IO–3 (aq) 0 0 s 2s
La expresión de la constante del producto de solubilidad será: K ps = [Ba2+] · [IO–3]2 = s · (2s)2 = 4s3.
En este caso, la constante del producto de solubilidad se expresará como: K ps = [Cl–] [Ag+]. Despejando, resulta: K ps [Cl–] = ––––––. [Ag+]
Sustituyendo los valores numéricos: 2 · 10–10 = 2 · 10 –7 M. [Cl–] = –––––––– 10–3
Sustituyendo, resulta: K ps = 4 · (7,2 · 10 –4)3 = 1,5 · 10–9. b) Si efectuáramos el cálculo de la solubilidad
en una solución de cloruro de bario, se tendría: [Ba2+] = s' + 0,3; [IO–3] = 2s' . Por tanto: K ps = 1,5 · 10–9 = (s' + 0,3) · (2s' )2. Efectuando la aproximación: s' + 0,3 Х 0,3 y resolviendo la ecuación anterior se obtiene: s' = 1,118 · 10 –5 M. Evidentemente, s' < s y, por
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3. La solubilidad del hidróxido de calcio es de 1,7 g/l a 298 K. Calcular a esta temperatura: a) La constante del producto de solubilidad. b) El pH de una disolución saturada de hidróxi-
do de calcio. RESOLUCIÓN: a) Para calcular la solubilidad molar es necesario conocer previamente la masa molar del hidróxido de calcio, que es de 74 g/mol. Así pues, se tiene: g 1 mol Ca(OH) l 74 g Todo el hidróxido se disociará, ya que las sales son electrolitos fuertes: 2 = 0,023 M s = 1,7 ––– · –––––––––––––––
Ca(OH)2 (s)
Inicio Equilibrio
s
–
→
Ca2+(eq) + 2 OH–(eq) 0 0 s 2s
La constante del producto de solubilidad vendrá expresada por: K ps = [Ca2+] [OH–]2 = 4s3. Sustituyendo los valores numéricos, resulta: K ps = 4 · (0,023) 3 = 4,87 · 10 –5. b) La concentración de hidróxido es: [OH –] = = 2s = 0,046 M. Así pues: pOH = –log 0,046 =
= 1,34. Por consiguiente: pH = 14 – pOH = 14 – 1,34 = 12,66.
= 0,065 M. El equilibrio de solubilidad que se produce es el siguiente: CH3—COO Ag(s) CH3—COO–(aq) + Ag+(aq)
La constante del producto de solubilidad s olubilidad viene dada por: K ps = [CH3—COO–] [Ag+]. Sustituyendo los valores numéricos, resulta: K ps = s2 = 0,06652 = 4,42 · 10 –3. b) Si consideramos que la
disolución es 0,2 M en nitrato de plata, hay que tener en cuenta que esta sal es un electrolito fuerte totalmente disociado: AgNO3 (aq) → Ag+(aq) + NO –3 (aq). Las concentraciones de los iones que intervienen en la solubilidad del acetato de plata serán: [CH3— COO–] = s'; [Ag +] = s' + 0,2. Por tanto, en este caso, la expresión de la constante del producto de solubilidad será: K ps = 4,42 · 10 –3 = = s' (s' + 0,2). Resolviendo la ecuación anterior, se obtiene: s' = 0,02 M. Se observa que s' < s y, por tanto, que la solubilidad del acetato de plata será menor en este caso, como consecuencia del efecto del ion común. 5. La solubilidad del hidróxido de hierro (II) en agua pura a 298 K es de 73,73 · 10 –5 g/l. Calcular a esta temperatura: a) El producto de
solubilidad del hidróxido de
hierro (II). 4. La solubilidad del acetato de plata a 298 K es de 11,1 g/l. Calcular a esta temperatura:
El producto de solubilidad del acetato de plata. b) La solubilidad del acetato de plata en una di-
b) Indicar si el hidróxido de
hierro (II) precipitará al añadir g de cloruro de hierro (II) a 1 litro de una disolución de pH = 6. 10–3
a)
RESOLUCIÓN: a) Para calcular la solubilidad molar es necesa-
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Fe2+(aq) + 2 OH–(aq) s Inicio – – Equilibrio 0 2s s siendo s la solubilidad molar del hidróxido de hierro (II). Todo Todo el hidróxido de hierro (II) disuelto se disocia, ya que las sales son electrolitos fuertes. Por tanto, se tendrá: K ps = [Fe2+] [OH–]2 = = (2s)2 · s = 4 s3. Sustituyendo los valores numéricos, resulta: Fe(OH)2 (s)
K ps = 4 s3 = 4 · (8,2 · 10 –6)3 = 2,2 · 10–15. b) El cloruro de hierro (II) es soluble
y se disocia totalmente: FeCl2 (aq) → Fe2+(aq) + 2 Cl–(aq). La concentración de [Fe 2+] disuelto será: –3 g FeCl 2+ 10 1 mol FeCl 1 mol Fe 2 2 2+ [Fe ] = –––––––––– · –––––––––– · ––––––––– =
l
127 g FeCl2 1 mol FeCl2 = 7,87 · 10–6 M
Como pH = 6, pOH = 14 – pH = 14 – 6 = 8. Ahora bien, por definición, pOH = –log [OH –], es decir, 8 = –log [OH –]. Por tanto: [OH –] = 10–8 M. Así pues, únicamente se obtendrá precipitado si Q > K ps. Q = [Fe2+] [OH–]2 = 7,87 · 10 –6 · (10–8)2 = = 7,87 · 10–22. En este caso, Q < K ps. Por consiguiente, no se producirá la precipitación del hidróxido de hierro (II). 6. A 25 °C, la constante del producto de solubilidad del hidróxido de manganeso (II) es 2,0 · 10–11. a) Calcular la solubilidad del hidróxido de manganeso (II) a dicha temperatura, expresada en mg/l. b) Calcular el pH de una disolución saturada de hidróxido de manganeso (II) a la temperatura indicada.
La constante del producto de solubilidad adoptará la expresión: K ps = [Mn2+] [OH–]2 = s · (2s)2 = 4 s3 = 2 · 10–11.
Así, resolviendo la ecuación anterior, se obtiene: s = 1,71 · 10–4 M. Para expresar la solubilidad en mg/l, procederemos del siguiente modo: 3 mg g mol 10 –4 s =1,71·10 –– ––––– ·88,9–––– ·–––––
– =15,2mg/l. l mol 1 g b) Teniendo en cuenta que [OH –] = 2s, siendo s la solubilidad molar, se obtiene la siguiente concentración de hidróxido: [OH –] = 3,42 · 10–4 M. De la definición de pOH = –log [OH –], sustituyendo, resulta: pOH = –log (3,42 · 10 –4) = 3,47. Por tanto: pH = 14 – pOH = 14 – 3,47 = 10,53 > 7. Es decir, se trata de un pH básico. 7. A 25 °C, la constante del producto de solubilidad del fluoruro de plomo (II) es 4 · 10 –8. a) Calcular la solubilidad del fluoruro de plomo (II) en agua pura a 25 °C, expresada en mg/l. b) Indicar si se formará precipitado de fluoruro de plomo (II) al mezclar, a 25 °C, 50 ml de disolución 0,02 M de nitrato de plomo (II) y 50 ml de disolución 0,01 M de fluoruro de potasio.
RESOLUCIÓN: a) En primer lugar, calcularemos la solubilidad molar a partir del equilibrio: Equilibrio
PbF2 (s)
Pb2+(aq) + 2 F–(aq) s
s
La constante del producto de solubilidad adop-
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g · –––––––– mol · 245,2 –––– 103 mg = = 2,154 · 10–3 –––– l mol 1g = 528,16 mg/l. b) Calcularemos la concentración de los iones en la mezcla: s
0,02 · 0,05 = 0,01 M; [Pb2+] = ––––––––– 0,1 0,01 · 0,05 = 0,005 M. [F–] = ––––––––– 0,1 A continuación determinaremos si se supera (precipita) o no (no precipita) el valor de la constante del producto de solubilidad. El cociente de reacción Q vale: Q = [Pb2+] [F–]2 = 0,01 · (0,005)2 = = 2,5 · 10–7 > K ps. Puesto que se supera el valor del producto de solubilidad, los iones no pueden estar en disolución con esta concentración y, por tanto, precipita el fluoruro de plomo (II). 8. El hidróxido de magnesio tiene una constante del producto de solubilidad a 25 °C de 1,0 · 10 –10. Calcular: a) La solubilidad del hidróxido de magnesio en mg/ml. b) El pH de una disolución saturada de hidróxido de magnesio.
RESOLUCIÓN: a) En primer lugar, calcularemos la solubilidad molar a partir del equilibrio: Mg(OH)2 (g) Equilibrio
Mg2+(aq) + 2 OH–(aq) s 2s
La constante del producto de solubilidad adoptará la expresión:
g mol l mol 1 l = 0,017 mg/ml. · –––––– 103 ml
mg g
s = 2,924 · 10–4 ––––– · 58,3 ––––– · 103 ––––– ·
b) Sabiendo que [OH –] = 2s, siendo s la solubi-
lidad molar, se obtiene la siguiente concentración de hidróxido: [OH –] = 5,848 · 10 –4 M. De la definición de pOH = –log [OH–], sustituyendo, resulta: pOH = –log (5,848 · 10 –4) = 3,23. Por tanto: pH = 14 – pOH = 14 – 3,23 = 10,77 > 7. Es decir, se trata de un pH básico. 9. A 25 °C, se disuelven en agua 0,148 g de yoduro de plomo (II) y se obtienen 250 ml de una solución saturada de esta sal. Calcular la constante del producto de solubilidad del yoduro de plomo a 25 °C.
RESOLUCIÓN: En 250 ml de disolución acuosa saturada, tenemos disueltos: 0,148 g PbI2 · –––––––––– 1 mol PbI2 = 1,28 · 10–3 M. ––––––––––– 0,25 l 461 g PbI2 El equilibrio de solubilidad que se establece es el siguiente: PbI2(s) Pb2+(aq) + 2 I–(aq)
Inicial Equilibrio
s
0
–
s
0 2s
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DESCRIPTI VA DE LOS PRINCIPALES DESCRIPTIV P RINCIPALES ELEMENTOS REPRESENTATIVOS 1. Calcular los kg de carbono que se pueden obtener a partir de 1.000 kg de carbonato de calcio, si el rendimiento del proceso es de un 30 %.
RESOLUCIÓN: El carbonato de calcio, como todos los carbonatos, a elevadas temperaturas se descompone del modo siguiente:
∆ CO + CaO. CaCO3 → 2 El proceso puede continuar:
∆ C+O. CO2 → 2 De este modo, la relación molar correspondiente a la totalidad del proceso es: 1 mol de carbonato de calcio Х 1 mol de dióxido de carbono Х 1 mol de carbono. Para calcular los gramos de carbono, necesitaremos determinar la masa molar del carbonato de calcio, que es de 100 g/mol. Así pues, se tiene: 1 mol CaCO3 · –––––––––––– 1 mol CO2 · 106 g CaCO3 · –––––––––––– 100 g CaCO3 1 mol CaCO3 1 mol C · ––––––––––––– 12 g carbono = 120.000 g C = · –––––––––– 1 mol CO2 1 mol carbono = 120 kg de carbono teóricos. Como el rendimiento es del 30 %, resulta: 30 kg C reales = 120 kg C teóricos · ––––––––––––––– 100 kg C teóricos = 36 kg de carbono.
RESOLUCIÓN: Un mol de cualquier gas ocupa un volumen de 22,4 litros en condiciones normales de presión y temperatura. Por tanto, se tiene: 1 mol H2 = 1,596 l H2 en c.n. · –––––––––––––– 22,4 l H2 en c.n. = 0,07125 moles de H 2. A continuación calcularemos la cantidad de hidrógeno que se produce de acuerdo con la ecuación igualada correspondiente al proceso: Zn + H2SO4
→
ZnSO4 + H2.
A partir de x gramos de Zn, se obtienen: 1 mol Z n 1 mol H 65,4 g Zn 1 mol Zn = 0,0153 x moles de H2.
2 = x g Zn · –––––––––– · ––––––––––
En el caso de la reacción con el aluminio se tiene: 2 Al + 3 H 2SO4
→
Al2(SO4)3 + 3 H2
A partir de (1,5 – x ) gramos de Al se obtienen: 1 mol Al · ––––––––––– 3 moles H2 = (1,5 – x ) g Al · ––––––––– 27 g A l 2 moles Al = 0,0833 – 0,0556 x moles de H2. Por tanto, el número total de moles de hidrógeno formados es: 0,0833 – 0,0556 x + 0,0153 x = = 0,0833 – 0,0403 x moles de H2. Ahora bien, los moles de hidrógeno procedentes de las dos reacciones han de coincidir con el número total de moles de hidrógeno obtenidos. Por tanto: 0,07125 = 0,0833 – 0,0403 x . Resolviendo esta
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3. Una muestra de 0,997 g de una mezcla de carbonato de sodio anhidro y de carbonato de potasio anhidro se trata con un ligero exceso de ácido clorhídrico diluido. La disolución resultante se evapora totalmente y el residuo obtenido pesa 1,090 g. Calcular la composición de la muestra original.
RESOLUCIÓN: En primer lugar, estableceremos las ecuaciones correspondientes a los procesos que tienen lugar: Na2CO3 + 2 HCl K2CO3 + 2 HCl
→ →
2 NaCl + CO2↑ + H2O 2 KCl + CO2↑ + H2O
Para calcular la cantidad de cloruro de sodio que se obtiene de la primera ecuación es necesario conocer las masas molares del carbonato de sodio y del cloruro de sodio, que son, respectivamente, 106 g/mol y 58,5 g/mol. Si consideramos que la muestra inicial contiene x gramos de carbonato de sodio, resulta: 1 mol Na CO 106 g Na2CO3
2 mol NaCl 1 mol Na2CO3
2 3 · –––––––––––– · x g Na2CO3 · ––––––––––––––
58,5 g NaCl = 1,104 x gramos NaCl. · ––––––––––– 1 mol NaCl Análogamente, para calcular la cantidad de cloruro de potasio que se obtiene de la segunda ecuación es necesario conocer las masas molares
formada por x + y = 0,997 (2). Resolviendo el sistema de ecuaciones constituido por (1) y (2) se obtiene: x = 0,552 g de carbonato de sodio; y = 0,445 g de carbonato de potasio. 4. Se calientan 5 g de carbonato de calcio y se obtienen 2 g de óxido de calcio y se libera dióxido de carbono. Calcular: a) El porcentaje en peso de carbonato de calcio que se ha transformado en óxido de calcio. b) El volumen de dióxido de carbono que se ha desprendido a 25 °C y 1 atm de presión.
RESOLUCIÓN: a) La ecuación correspondiente al proceso de descomposición es la siguiente: CaCO3 (s)
∆
→
CaO(s) + CO2 (g).
Para efectuar los cálculos estequiométricos necesitamos conocer previamente las masas molares del carbonato de calcio y del óxido de calcio, que son, respectivamente, 100 g/mol y 56 g/mol. Calcularemos los gramos de óxido de calcio que se obtendrían a partir de los 5 gramos de carbonato de calcio: 1 mol CaCO3 · –––––––––––– 1 mol CaO · 5 g CaCO3 · –––––––––––– 100 g CaCO3 1 mol CaCO3 56 g CaO
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1 mol CaO · –––––––––– 1 mol CO2 = 2 g CaO · –––––––––– 56 g CaO 1 mol CaO = 0,036 moles de CO2.
A continuación, calcularemos los moles de hidrógeno que corresponden a ambas reacciones a partir de x gramos de Mg y de (4 – x ) gramos de Al:
Aplicando la ecuación de estado de los gases ideales:
2 = x g Mg · –––––––––– · –––––––––
n R T P V = n R T , despejando resulta: V = ––––––. P
Sustituyendo los valores numéricos: J · 298 K 0,036 mol · 8,31 ––––––– K · mol V = ––––––––––––––––––––––––––––– = 1,01 · 105 Pa 1000 litros = 0,883 litros. = 8,83 · 10–4 m3 · –––––––––– 1 m3 5. En la construcción aeronáutica se utiliza una aleación de magnesio y aluminio denominada magnalio. Al reaccionar 4 g de dicha aleación con un exceso de ácido clorhídrico, se obtuvieron 4,72 litros de hidrógeno, medidos a 1 atm y a 0 °C. Calcular el porcentaje de magnesio presente en el magnalio.
RESOLUCIÓN: En primer lugar, estableceremos las dos ecuaciones que tienen lugar: Mg + 2 HCl → MgCl2 + H2 2 Al + 6 HCl → 2 AlCl + 3 H
1 mol Mg 1 mol H 24,3 g Mg 1 mol Mg = 0,0412 x moles H2
1 mol Al · ––––––––––– 3 moles H2 = (4 – x ) g Al · ––––––––– 27 g Al 2 moles Al = (0,2222 – 0,0556 x )moles H2 Esta cantidad de moles ha de coincidir con los moles totales de hidrógeno formados. Así, se tiene: 0,208 = 0,0412 x + (0,2222 – 0,0556 x ). ). Resolviendo la ecuación anterior se obtiene: x = 1 gramo de Mg; (4 – x ) = 4 – 1 = 3 gramos de Al. Por tanto, el porcentaje de magnesio en el magnalio será: 1 g Mg · 100 = 25 % de Mg. ––––––– 4g 6. El carbonato de magnesio constituye el mineral denominado magnesita. Al calentarlo a temperatura elevada se descompone en óxido de magnesio y dióxido de carbono. Calcular la masa de magnesita, que contiene un 90 % de carbonato de magnesio, que se necesita para obtener 10,0 g de óxido de magnesio, si el rendimiento del proceso es del 80 %.
RESOLUCIÓN:
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METALURGIA 1. a) Calcular la masa de cinc que se puede obtener a partir de una tonelada de blenda de un 40 % de riqueza. b) Calcular el volumen de dióxido de azufre en condiciones normales que se desprende en el proceso de tostación.
RESOLUCIÓN: a) La reacción que se produce en el proceso de tostación es la siguiente: 2 ZnS + 3 O2
→
2 SO2 + 2 ZnO
La masa de sulfuro de cinc que contiene 1 tonelada de blenda es: 1.000 kg blenda · 1 tonelada blenda · –––––––––––––––– 1 tonelada blenda 40 kg ZnS = 400 kg ZnS · ––––––––––––– 100 kg blenda El cinc se obtiene por reducción del óxido de cinc producido en el proceso de tostación: ZnO + CO
→
2. Se prepararon 25 g de manganeso por aluminotermia de una pirolusita (dióxido de manganeso) del 50 % de riqueza. Calcular la cantidad de pirolusita empleada en la aluminotermia.
RESOLUCIÓN: En la aluminotermia de la pirolusita tiene lugar la siguiente reacción de reducción: 3 MnO2 + 4 Al
→
3 Mn + 2 Al 2O3
Para obtener 25 gramos de manganeso, se precisaría: 1 mol Mn · –––––––––––––– 3 moles MnO2 · 25 g Mn · –––––––––– 54,9 g Mn 3 moles Mn 86,9 g pirolusita = 39,57 g de pirolusita pura. · ––––––––––––––– 1 mol MnO2 Por tanto, la cantidad de pirolusita impura necesaria es: impu pura ra = 100 g im 39,57 g pura · –––––––––––– 50 g pura = 79, 14 g de pirolusita impura.
CO2 + Zn
Así pues, la cantidad de Zn obtenida a partir de 1 tonelada de blenda es: 1.000 g ZnS · ––––––––––– 1 mol ZnS · 400 kg ZnS · ––––––––––– 1k Z S 97 37 Z S
3. Calcular los gramos de nitrato de plata necesarios para fabricar una moneda de plata de 25 g cuya ley es del 90 %.
RESOLUCIÓN:
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4. Un mineral de galena con ganga de cuarzo tiene una densidad de 4,9 g/cm 3, siendo la densidad de la galena 7,58 g/cm 3 y la del cuarzo 2,65 g/cm3. Calcular la cantidad de plomo que puede obtenerse teóricamente a partir de una tonelada de dicho mineral.
5. Para determinar la riqueza de una muestra de cinc impuro se toman 50 g y se tratan con una disolución clorhídrica de densidad 1,18 g/cm 3 y 35 % de riqueza en peso y se gastan 129 ml de dicha disolución. Calcular el porcentaje de cinc contenido en la muestra.
RESOLUCIÓN: Primero calcularemos la cantidad de galena presente en el mineral, hallando la composición centesimal. Supongamos que el mineral contiene un x % de galena y, por tanto, un (100 – x )% )% de cuarzo. La densidad del mineral vendrá dada por la expresión:
RESOLUCIÓN: En primer lugar, calcularemos la cantidad de ácido clorhídrico que ha reaccionado con la muestra de cinc impuro:
x Densidad Densida d del miner mineral al = densidad galena · –––– +
100
100 – x + densidad cuarzo · –––––––. 100 Sustituyendo los valores numéricos del enuncia x 100 – x do, resulta: 4,9 = 7,58 · –––– + 2,65 · –––––––. 100 100 225 = 45,641% de Despejando, se obtiene: x = ––––– 4,93 galena. Por tanto, la galena contenida en una tonelada de mineral será:
1,18 g disolución · –––––––––––––– 35 g HCl 129 ml · ––––––––––––––– · ml 100 g disolución 1 mol HCl = 1,4616 moles HCl. · ––––––––––– 36,45 g HCl La reacción que tiene lugar entre el cinc y el ácido clorhídrico es la siguiente: Zn + 2 HCl
→
ZnCl2 + H2
Así pues, a partir de la relación estequiométrica anterior resulta:
kg mineral· ––––––––––––––– 45,641 kg galena = 1 tonelada · 1.000 ––––––––––––– 1 tonelada 100 kg mineral = 456,4 kg de galena.
1 mol Zn · –––––––––– 65,37 g Zn = 1,4616 moles HCl · ––––––––––– 2 moles HCl 1 mol Zn = 47,77 g Zn.
El plomo se obtiene a partir de la galena por tostación y posterior reducción, de acuerdo con
Por consiguiente, el porcentaje de cinc contenido en la muestra será:
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1.000 kg calcita · 1 tonelada de calcita · –––––––––––––––––– 1 tonelada de calcita 80 kg CaCO3 = · ––––––––––––– 100 kg calcita = 800 kg de carbonato de calcio. La reacción de calcinación del carbonato de calcio es la siguiente: CaCO3
∆ CO + CaO. 2
→
La cantidad teórica de óxido de calcio que se obtendría en el proceso sería: 1.000 g CaCO3 · ––––––––––– 1 mol CaCO3 · 800 kg CaCO3 · ––––––––––––– 1 kg CaCO3 100 g CaCO3 1 mol CaO · –––––––––– 56 g CaO · ––––––––––– 1 kg CaO = · –––––––––––– 1 mol CaCO3 1 mol CaO 1.000 g CaO = 448 kg CaO. Pero, como el rendimiento del proceso es del 90 %, la cantidad de óxido de calcio que se obtiene realmente es:
En el primer caso, se tiene: g · –––––––––– 1 mol M 1 mol––––– H2 · ––––––– –––– ––––– x g Mg · ––––––––––– 24,31 g Mg 1 mol Mg 22,4 l H2 (c.n.) · –––––––––––––––– 1.000 ml H2 (c.n.) = · –––––––––––––– 1 mol H2 1 l H2 (c.n.) = 921,43 x ml de H2 (c.n.). En el segundo caso resulta: 1 mol Al · ––––––––––– 3 moles H2 · (0,5 – x ) g Al · ––––––––––– 26,98 g Al 2 moles Al 22,4 l H2 (c.n.) · –––––––––––––––– 1.000 ml H2 (c.n.) = · –––––––––––––– 1 mol H2 1 l H2 (c.n.) = (622,68 – 1.245,37 x ) ml de H2 (c.n.). Ahora bien, la suma de los volúmenes de hidrógeno desprendidos en ambos procesos debe coincidir con el volumen total de hidrógeno desprendido en el ataque, es decir, 602,4 ml. Por tanto, se tiene: 921,43 x + (622,68 – 1.245,37 x ) = 602,4. Resolviendo la ecuación precedente se obtiene: x = 0,063 g Mg. Por consiguiente:
448 kg · 90 % = 403,2 kg de óxido de calcio = = 403 200 g de CaO.
(0,5 – x ) = 0,5 – 0,063 = 0,437 g Al.
7. Al tratar 0,5 g de una aleación de magnesio y aluminio con ácido clorhídrico se desprenden 602,4 ml de hidrógeno medidos en condiciones
8. Cuando un trozo de hierro de 10 g se expone a la intemperie su masa aumenta en 0,5 g. Suponiendo que únicamente se haya formado óxido
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1 mol Fe 2 moles Fe O 55,85 g Fe 4 moles Fe 159,7 g Fe2O3 = 1,4297 x g Fe O . · –––––––––––––– 2 3 1 mol Fe2O3
2 3 · x g Fe · –––––––––– –––––––––– · –––––––––––––
Al finalizar el proceso, la suma de las masas del óxido de hierro (III) formado y del hierro que no ha reaccionado deberá coincidir con la masa final, que es de 10,5 gramos. Así pues, tendremos: 1,4297 x + 10 – x = 10,5. Resolviendo la ecuación precedente, se obtiene: x = 1,164 gramos de hierro que se han oxidado. Por consiguiente: (10 – x ) = 10 – 1,164 = 8,836 gramos es la masa de hierro que no ha reaccionado.
MÉTODOS DE LA QUÍMICA ANALÍTICA 1. Un ácido desconocido se valora frente al carbonato de sodio puro y anhidro. 0,2360 g de carbonato de sodio precisan para su total neutralización 22,15 ml del ácido. Calcular la normali-
carbonato de sodio = 1 · (2 · 1) = 2 equivalentesgramo. El resultado que se obtiene es el mismo si se multiplica el número de moles de carbonato de sodio (1) por el producto del número de aniones carbonato (1) por la carga de cada anión carbonato (2). En efecto, se obtiene que 1 mol de carbonato de sodio = 1 · (1 · 2) = 2 equivalentes-gramo. Por tanto, se tiene que 2,226 · 10 –3 moles de 2 equivalentes-g Na2CO3 = carbonato de sodio · ––––––––––––––––––––– 1 mol Na2CO3 = 0,004452 equivalentes-gramo de carbonato de sodio. Ahora bien, todos los procesos químicos tienen lugar reaccionando y formándose el mismo número de equivalentes-gramo de todas las sustancias que intervienen en el mismo. Por tanto, habrán reaccionado también 0,004452 equivalentes-gramo del ácido desconocido HA. Por definición, la normalidad N de una disolución es el cociente entre el número de equivalentes-gramo del soluto y el número de litros de disolución. En este caso, se tendrá:
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1 mol NaOH = 0,025 moles NaOH. 1 g NaOH · –––––––––––– 40 g NaOH En el caso de un hidróxido, el número de equivalentes-gramo valentes-gr amo se calcula multiplicando el número de moles por el número de grupos OH que posee. Es decir, en el caso que nos ocupa, 1 mol de hidróxido de sodio es igual a 1 equivalente-gramo, equivalente-gramo, ya que hay que multiplicar el número de moles de hidróxido de sodio (1) por el número de grupos OH (1). Así pues, 1 mol de hidróxido de sodio = = 1 · 1 = 1 equivalente-gramo. Por tanto, se tiene que 0,025 moles de NaOH = 0,025 equivalentesgramo de NaOH. Como en todos los procesos químicos interviene el mismo número de equivalentes-gramo de todas las sustancias, se habrán gastado también 0,025 equivalentes-gramo del ácido HA. De la definición de normalidad, se tiene:
mo, ya que hay que multiplicar el número de moles de bicarbonato de sodio (1) por el producto del número de cationes Na+ (1) por la carga de cada catión Na+ (1). Así pues, 1 mol de carbonato de sodio = 1 · (1 (1 · 1) = 1 equivalente-gram equivalente-gramo. o. El resultado que se obtiene es el mismo si se multiplica el número de moles de bicarbonato de sodio (1) por el producto del número de aniones bicarbonato (1) por la carga de cada anión bicarbonato (1). Por tanto, se tiene que 0,0119 moles de bicarbonato de sodio son también 0,0119 equivalentes-gramo de bicarbonato de sodio. Como en todos los procesos químicos interviene el mismo número de equivalentes-gramo de todas las sustancias, se habrán gastado también 0,0119 equivalentes-gramo del ácido HA. De la definición de normalidad, se tiene:
número de equivalentes-gramo ––––––––––––––––––––––––––––. ––. N = –––––––––––––––––––––––––– V (litros)
número de equivalentes-gramo N = –––––––––––––––––––––––––––. V (litro)
Despejando: V · N = número de equivalentesgramo. Sustituyendo los datos numéricos, resulta: V · 0,5 = 0,025. Despejando:
Despejando: V · N = número de equivalentes-gramo.
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das las sustancias, se habrán gastado también 0,01887 equivalentes-gramo del ácido HA. De la definición de normalidad, se tiene: número de equivalentes-gramo –––––––––––––––––––––––––––. . N = ––––––––––––––––––––––––––– V (litros)
Despejando: V · N = número de equivalentesgramo. Sustituyendo los datos numéricos, resulta: V · 0,5 = 0,01887. Despejando: 0,01887 0,5
V = –––––––– = 0,03774 l = 37,74 ml.
3. 20 ml de una disolución que contiene potasa cáustica y carbonato de potasio se valoran en frío con un ácido 0,2 N, factor 0,992. Empleando fenolftaleína como indicador, se gastan hasta decoloración del líquido 27,40 ml del ácido. Se agrega ahora rojo de metilo y el líquido se vuelve amarillo y, y, al continuar valorando hasta débil coloración rojiza se gastan 5,85 ml del ácido. Calcular la concentración de la disolución en hi-
co) a color rojizo (medio ácido). La reacción que se produce en este caso es la siguiente: KHCO3 + HA
→
KA + CO2 + H2O
De la definición de normalidad, se deduce que V · N = número de equivalentes-gramo de ácido. Sustituyendo los valores numéricos, se obtiene para la primera valoración: Equivalentes-gramo de ácido = 27,40 · 10–3 · 0,2 · 0,992 = 0,005436. Para la segunda valoración, resulta: Equivalentes-gramo de ácido = 5,85 · 10 –3 · 0,2 · 0,992 = 0,001162. Ahora bien, como el número de equivalentesgramo ha de ser el mismo en cualquier reacción, en la segunda valoración también habrán reaccionado 0,001162 equivalentes-gramo de bicarbonato de potasio, que a su vez, provienen de 0,001162 equivalentes-gramo de carbonato de potasio que han reaccionado en la primera valoración. Por tanto, estos 0,001162 equivalentesgramo de carbonato de potasio habrán reaccionado con 0,001162 equivalentes-gramo de ácido en la primera valoración y, por tanto, el ácido total
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0,001162 moles K2CO3 · –––––––––––– 138 g K2CO3 · –––––––––––––––––––– 20 ml 1 mol K2CO3 1.000 mol = 8,02 g/l · ––––––––– 1l Hidróxido de potasio: 0,004274 moles KOH · –––––––––– 56 g KOH · –––––––––––––––––––– 20 ml 1 mol KOH 1.000 ml = 12 g/l · ––––––––– 1l 4. Para determinar el calcio presente en un agua mineral, se precipita como oxalato de calcio monohidrato, mediante oxalato de amonio a ebullición en presencia de cloruro de amonio. El oxalato de calcio se filtra, se lava cuidadosamente, se disuelve después en ácido sulfúrico y se valora con permanganato 0,1 N, factor 1,012. Partiendo de 100 ml de agua mineral se gastan 13,95 ml de permanganato. Calcular el contenido del agua en calcio, determinándolo como ion
de calcio y, por tanto, con los de Ca 2+. Ahora bien, en el caso de un catión, el número de equivalentes-gramo coincide con el número de moles multiplicado por la carga del catión. Es decir, en el caso del catión Ca2+, 1 mol = 1 · 2 = 2 equivalentes-gramo. Así pues, efectuando los cálculos estequiométricos correspondientes se obtiene: 0,0014117 equivalentes-gramo 1 mol Ca2+ ––– · –––––––––– 40 g Ca2+ = Ca2+ · –––––––––––––––––––––– ––––––––––––––––––– 2 equivalentes-gramo Ca 2+ 1 mol Ca2+ = 0,0283 gramos de Ca2+. Pero como esta cantidad de Ca 2+ está disuelta en 100 ml, la concentración será: 2+ g 0,0283 Ca 1.000 ml = 0,283 g/l. ––––––––––––– · –––––––– 100 ml 1l Si se quiere expresar la concentración en forma de óxido de calcio se procederá del modo siguiente: 2+ 2+ g Ca 1 mol Ca 1 mol CaO · 0,283 ––––––– · –––––––––– · ––––––––––– l 40 g Ca2+ 1 mol Ca2+
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14 H+ + K2Cr2O7 + 6 Fe2+ → 3+ + 6 Fe3+ + 7 H O + 2 K+ → 2 Cr 2 A partir de 1 g de mineral tendremos: 1 g · 1% = = 0,01 g de hierro. Calculemos el número de moles de Fe3+ en que se transforma: 3+ 1 m o l F e 2 m o l e s F e 0,01 g Fe · ––––––––– · ––––––––––– = 55,8 g Fe 2 moles Fe
= 1,792 · 10–4 moles Fe3+. A continuación vamos a determinar el número de equivalentes-gramo de Fe(II) que se obtienen en el proceso de reducción de Fe(III) a Fe(II). En un proceso redox, el número de equivalentesgramo de una sustancia coincide con el número de moles por el número de electrones transferidos en el proceso. En el caso de la oxidación del Fe(II) a Fe(III) se tiene que 1 mol = 1 · 1 = 1 equivalente-gramo, ya que únicamente se transfiere 1 electrón para pasar de Fe(II) a Fe(III). Por tanto, 1,792 · 10–4 moles Fe3+ = 1,792 · 10–4 moles Fe2+ = 1,792 · 10–4 equivalentes-gramo de
6. Se analiza una mezcla de arsenito y arseniato. Se disuelven 0,500 g de la mezcla y en medio neutro se valora con yodo 0,1 N, necesitándose 20,95 ml. La disolución resultante se acidifica fuertemente, se añade yoduro de potasio y el yodo liberado se valora con tiosulfato 0,1 N. Se gastan 48,35 ml de tiosulfato. Calcular la proporción de arsenito y arseniato en la muestra, calculados como As2O3 y As2O5, respectiv respectivamente. amente.
RESOLUCIÓN: En la primera valoración, el yodo ha oxidado el arsenito a arseniato de acuerdo con la siguiente reacción: I2 + AsO3–3 + H2O
→
2 I– + AsO3–4 + 2 H+
La disolución resultante contiene únicamente arseniato. Al acidificar fuertemente y añadir yoduro de potasio, el arseniato oxida al yoduro a yodo de acuerdo con la siguiente reacción: AsO3–4 + 2 I– + 2 H+
→
I2 + AsO3–3 + H2O
El yodo que se libera reacciona con el tiosulfa-
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Así pues, el porcentaje de As2O3 en la muestra es: 0,1036 g · 100 = 20,7% As O . ––––––––– 2 3 0,5 g A continuación calcularemos el número de equivalentes-gramo de tiosulfato que han reaccionado en el segundo proceso: Número de equivalentes-gramo de tiosulfato = V · N = = 48,35 · 10–3 · 0,1 = 4,835 · 10 –3. Éste será también el número de equivalentes-gramo de yodo que habrán reaccionado y, por tanto, de arseniato total. Ahora bien, si de este número de equivalentes-gramo de arseniato descontamos los equivalentes-gramo de arsenito determinados en el primer proceso, que posteriormente se convier-
co isómero con una cadena de 5 átomos de carbono es el n-pentano: CH3—CH2—CH2—CH2—CH3. Con una cadena principal de 4 átomos de carbono tan sólo puede existir el metilbutano: CH3—CH—CH2—CH3 |
CH3 ya que no es posible tener isómeros de posición, puesto que serían el mismo compuesto empezando a numerar la cadena por el otro extremo. Por último, con una cadena principal de 3 átomos de carbono, solamente puede existir el 2,2-dimetilpropano: CH3 |
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5. Como sólo hay un átomo de carbono, únicamente puede ser: H |
H—C—Br |
I b) 1. El grupo funcional es el grupo alcohol: —OH. \ 2. El grupo funcional es el grupo aldehído: C O ⁄ 3. El grupo funcional es el grupo ácido carboxílico: O ⁄ ⁄ —C \ O ϭ
RESOLUCIÓN: El compuesto es un hidrocarburo monoinsaturado con 3 átomos de carbono. Por tanto, se puede pensar «a priori» que existan isómeros lineales y cíclicos. Por lo que respecta a los isómeros lineales, únicamente pueden existir los que tengan 3 átomos de carbono en la cadena principal, ya que es imposible tener cadenas de 2 átomos de carbono con ramificaciones, puesto que la cadena ramificada sería más larga que la principal. Como los posibles isómeros lineales de 3 átomos de carbono deben poseer un doble enlace, resulta igual colocarlo en una posición que en otra, ya que los localizadores serán los mismos empezando a contar desde el otro extremo de la cadena principal. Por
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CH2 ⁄ \ f ) H2C—CH2 + Br2
→
BrCH2—CH2—CH2Br
O ⁄ ⁄ ⁄⁄ ∆ g) 3 CH3—C + 2 MnO–4 + OH– → \ H O ⁄⁄ ⁄ ⁄ ∆ — C + 2 MnO2 + 2 H2O → 3 CH3 \ O– h) HCCl3 + OH– → :CCl2 + H2O + Cl– RESOLUCIÓN: a) Se trata de una reacción de bromación, es decir, de una reacción en la cual se produce la
do carboxílico mientras que el permanganato se reduce a manganeso (IV). h) Se trata de una reacción de eliminación, concretamente de una deshidrohalogenación, ya que el compuesto inicial pierde HCl para conv converertirse en un radical. 6. De las siguientes especies indicar las que se comportan como nucleófilas y electrófilas: a) Cl–; b) H2O; c) H+; d ) AlBr3; e) CH3—OH; f ) BeCl2; g) Br+; h) Cr3+; i) SnCl4; j) NO+2; k ) H2C O; l ) CH3—CN; m) H2; n) CH4; o) H2C CH—CH3. ϭ
ϭ
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l)
Se trata de una especie que es a la vez nucleófila y electrófila, ya que el carbono es electrófilo pero el nitrógeno es nucleófilo. m) Se trata de una especie que no es ni nucleófila ni electrófila, ya que no tiene tendencia ni a dar ni a recibir electrones. n) Se trata de una especie que no es ni nucleófila ni electrófila, ya que no tiene tendencia ni a dar ni a recibir electrones. o) Se trata de una especie nucleófila, ya que el doble enlace puede reaccionar con una especie de baja densidad electrónica. 7. Clasificar las siguientes reacciones como de
d )
Se trata de una reacción redox de desplazamiento en que el hidruro se oxida a hidrógeno y el bromo se reduce a bromuro. e) Se trata de un proceso redox de adición de hidrógeno al doble enlace de un alqueno. El alqueno se reduce a alcano y el hidrógeno se oxida a H +. f ) Se trata de una reacción de sustitución en un anillo bencénico, concretamente de una nitración, en la cual un grupo nitro sustituye a un hidrógeno. g) Se trata de una reacción de eliminación, en la que un ácido carboxílico al ser calentado a temperaturas elevadas elimina una molécula de agua de su estructura con desprendimiento de monóxido de carbono. h) Se trata de una reacción de adición de agua
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H2C—CH—CH—CH3 |
|
→
|
H Br Br H CH—CH2—CH3 + → H 2C + CH3—CH CH—CH3 De los dos productos se obtendrá preferentemente el 2-buteno ya que por la regla de Saycheff, el orden de preferencia para eliminar los H en las reacciones de deshidrohalogenación es: terciario > secundario > primario y, en este caso, el 2-buteno se forma a partir de la eliminación de un H situado en un átomo de carbono secundario, mientras que el 1-buteno se forma a partir de la eliminación de un H situado en un átomo de carbono primario. Como el 2-buteno ϭ
ϭ
De los dos productos se obtendrá preferentemente CH3 |
CH3—C
CH—CH2—CH3
ϭ
ya que por la regla de Saycheff, el orden de preferencia para eliminar los H en las reacciones de desdes hidrohalogenación es: terciario > secundario > primario y, en este caso, el CH3 |
CH3—C CH—CH2—CH3 se forma a partir de la eliminación de un H situado en un átomo de carbono secundario, mientras que el CH3 ϭ
|
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f )
En este caso, el átomo de Br tiene tres átomos de carbono adyacentes, por lo que se obtendrán los tres productos de reacción siguientes:
|
H2CH CH3—CH—C—C—(CH3)2 |
|
a) CH3—C—C—CH3;
|
|
2. Indicar las fórmulas de los alquenos formados por la deshidratación de los siguientes alcoholes: CH3 H CH3
|
→
|
|
OH CH3
|
b) CH3—C—CH2OH |
CH3
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se forma a partir de la eliminación de un H situado en un átomo de carbono primario. b) En este caso, el alcohol neopentílico no puede deshidratarse directamente porque no hay átomos de H en el carbono adyacente. El catión neopentilo formado inicialmente es primario y, por tanto, experimenta una reordenación para convertirse en un catión terciario mucho más estable. Finalmente
CH3 |
d ) CH3—CBr—CH3 + CH3—COO–
→
RESOLUCIÓN: a) Se trata de una reacción de deshidrohalogenación, en la que se forma preferentemente
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CH3 |
d ) m-xileno + CH 3—C—Cl |
AlCl3 → 100 °C
CH3 RESOLUCIÓN: a) La secuencia de reacciones que se producen es la siguiente:
CH3 |
p—CH3—C6H4—C—CH2—CH3 |
CH3 c) La secuencia de reacciones que se produce es la que se indica a continuación: AlCl3 CH CH CH CH Cl →
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bien un triple enlace (alquino). El hecho de que origine un producto carbonílico al ser tratado con una solución acuosa de H2SO4 + HgSO4 nos indica que se trata de un alquino, que adiciona agua en su estructura y, posteriormente, transforma el enol obtenido en un compuesto carbonílico. El hecho de que el compuesto carbonílico formado no reaccione con solución amoniacal de nitrato de plata
cir, puede tratarse de un dieno o bien de un alquino. El hecho de que D reaccione con CuCl amoniacal, indica que D es un alquino que posee el triple enlace en el extremo de la cadena carbonada por lo que su átomo de hidrógeno situado en dicha posición tiene características ácidas. Así pues, el compuesto D únicamente puede ser HC C—CH2—CH3, C debe ser forzosamente BrCH CHBr CH ϵ
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Por tanto, el alcohol formado es: CH3 ⁄ —CHOH—CH \ CH3 c) La secuencia de reacciones que tiene lugar es la siguiente:
3. Los siguientes reactivos de Grignard y aldehídos o cetonas reaccionan y los productos resultantes son hidrolizados. Indicar el nombre del alcohol obtenido en cada caso: a) benzaldehído y C2H5MgBr; b) acetaldehído y bromuro de bencilmagnesio; c) acetona y bromuro de bencilmagnesio; d ) formaldehído y bromuro de ciclohexilmagnesio; ) acetofenona y bromuro de etil-
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Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
The world's largest digital library
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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The world's largest digital library
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