TALLER: GRAVIMETRIA Y APLICACIONES DE LAS TITULACIONES DE NEUTRALIZACION.
INFORME PRESENTADO POR: GEMALY ALVAREZ BAJAIRE 1, JOSÉ CASTRO RODRÍGUEZ 1, LADY JATTIN TORES1, MABEL ROCIO OLAYA PEREZ1, MELISSA PINEDA MERCADO 1. 1.
Estudiantes de la Universidad Universidad de Cartagena, Programa Programa de Ingeniería Ingeniería Química.
TRABAJO PRESENTADO A: CANDELARIA TEJADA TOVAR
UNIVERSIDAD DE CARTAGENA FACULTAD DE INGENIERÍAS PROGRAMA DE INGENIERÍA QUÍMICA LABORATORIO DE QUIMICA ANALITICA
CARTAGENA DE INDIAS D. T. y C. MARTES 17 DE ABRIL, 2012
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
PROBLEMAS DEL IBARZ-GRAVIMETRIA. IBARZ-GRAVIMETRIA. 10. En el análisis de una muestra de giobertita se pesan 2.8160g de mineral, que se disuelven en acido clorhídrico diluido; el liquido se lleva a un volumen de 250 cm 3. Se toman 50 cm3 de este líquido, se añade amoniaco en exceso y se precipita con fosfato sódico. El precipitado obtenido se filtra, lava, seca y calcina en cuyo proceso el fosfato magnésico formado se transforma en pirofosfato. Se obtienen 0.622g de Mg 2P2O7. Calcular el contenido en magnesio de aquel mineral y su riqueza en carbonato magnésico, MgCO 3. R/
1 mol de Mg PO + = 250 cm cm × 0.622g50decmMg PO × 222.5579 g Mg P O
+ 2 mol Mg g de Mg + = 0.6792 + × 1 mol Mg P O × 24.3050 0.6 792g g de Mg M g + 1 mol Mg 0.6792g de Mg + x 100 + 10 0 = 24.13 24. 13% % de Mg 2.8160g de giobertita
+ 1 mol de Mg de MgCO x 84.305 g de MgCO + = 0.6792 0.6792gg de Mg Mg X 24.3050 g de Mg + X 1 1mol mol Mg+ 1 mol MgCO = 2.356g 2.356g de de MgCO MgCO 2.356g de MgCO x 100 83.66% 6% de MgCO MgCO 2.8160g de giobertita 100 = 83.6 11. En el análisis de una mezcla salina se encuentra que en 1g de la misma hay 0.2014g de ion Br -, 0.529g de ion NO 3-, 0.171g de ion Ca ++ y el resto, 0.0986g, de ion K +. Calcular la composición de aquella mezcla. R/ 0.2014g Br - + 0.529g NO 3- + 0.171g Ca ++ + 0.0986g K + = 1g mezcla salina
1 mol de Br − = 0.00252 − 0.00 252 mol de Br − 79.909g de+Br 1 mol de K = 0.00252 + mol K + = 0.09 0.0986 86gg K + X 39.102 0.00 252 mol de K + g de K mol Br − = 0.201 0.2014g 4g Br−X
Son equimolares lo que hace sospechar que estos iones están juntos como KBr, es decir,
0.00252 mol de KBr.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
1 mol de Ca++ = 0.0043 ++ 0.00 43 mol m ol de d e Ca ++ 40,08g de Ca Los moles de NO − son el doble que los moles de Ca++ , lo que hace sospechar que estos iones están juntos como Ca ( NO 2 es decir, 0.0042 mol de Ca (NO 2 KBr = 0.3g g KBr KBr = 0.002 0.00252 52 mol mol KBr KBr x 119.002g KBr 1 mol de KBr 0.3g KBr Ca NO = 0.70g g Ca NO = 0.0043 0.0043 mol mol Ca Ca NO NO x 164.091g NO 1 mol de Ca NO 0.70g Ca NO
mol Ca++ =
0.171g 171g Ca Ca++ X
Finalmente
0.299g KBr X 10 KBr 1 g muestra salina 100 = 30% KB 0.70g Ca NO X 10 Ca NO 1 g muestra salina 100 = 70% Ca 12. Una disolución acida contiene, además del ion H +, los iones K +, Cl- y SO4--. En la determinación de la acidez, 50 cm 3, de disolución se neutralizan con 0,2 g de sosa caustica. El ion K + se precipita en condiciones adecuadas como cloroplatinato potásico,K 2Cl6Pt; a partir de 10 cm 3 de disolución se obtienen 0,730 g de K 2Cl6Pt . En la precipitación del ion Cl -, de 25 cm 3 de disolución se obtienen 0,717 g de AgCl y al precipitar el ion SO4- como sulfato barico se forman 1,167 g de BaSO 4 a partir de 50 cm3 . Calcular la composición de la disolución y cual a podido ser la forma de prepararse.
0,717 g AgCl ×
1 mol de AgCl 143,321 g AgCl
0,005 moles de Cl 25
cm3
-
×
1000 cm3 1L
×
1 mol de Cl
-
1 mol de AgCl
= 0,2 de
moles L
=0,005 moles de Cl
Cl
-
-
−− 1 1 1,167 × 233,3976 × 1 = 0,005 0,005 − − --
0,005 moles de SO 4 50
cm3
×
1000 cm3 1L
= 0,1 0,1 de
moles L
--
SO4
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
0,003 moles de K + 10
cm3
0,2 g NaOH ×
1L
= 0,3 0,3 de de
1 mol de NaOH 39,99707 g de NaOH
0,005 moles de H + 50
×
1000 cm3
cm3
×
1000 cm3 1L
×
moles L
+
K
1 mol de H + 1 mol de NaOH
= 0,2 0,2 de de
moles L
Ag: 107.868 g O: 15.9994 g Cl: 35.453 g K: 39.098 g Ba: 137.34 g Cl: 35.453 g S: 32.06 g Pt: 195.09 g Tabla-Pesos moleculares (g/mol)
=0,005 moles de H
+
+
H
Na: 22.98977 g H: 1.0079 g
Por otra parte parte la disolución disolución se pudo preparar a parir de 0,1 mol de HCl/ L; L; 0,1 mol de KCl,/L y 0,1 mol K 2SO4 /L o también a parir de 0,2 moles de KCl /L y O,2 moles de KHSO4 /L.
14. 0,852g de una aleación aleación de aluminio y cobre cobre se disuelven en ácido ácido nítrico, el líquido se evapora a sequedad y la mezcla de nitratos se calcina, se obtienen 1,566g de una mezcla de óxidos de aluminio (Al 2O3) y de cobre (CuO). En un segundo ensayo a partir también de 0,852g de aleación se obtiene 1,567g de la mezcla de óxidos. Calcular de ambas determinaciones la composición de aquella aleación. Solución: Al Cu O Al2O3 CuO 26,981 63,546 15,999 101,959 79,536 Tabla-Pesos moleculares (g/mol) Suponemos que:
= =
PRIMER ENSAYO: La aleación de plata y cobre tienen una masa de 0,852g entonces:
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
1 mol de Cu × 1 mol de CuO × 79,536g CuO ? gCuO gCuO = xg Cu × 63,546 g Cu 1 mol de Cu 1 mol de CuO = 1,251 1,251gg CuO 3 mol de Al × 1 mol de AlO × 101,959g AlO ?gAlO =yg Al× 126,981g Al 2 mol de Al 1 mol de AlO = 1,88 ,889g AlO 4 Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCuOx gAlOy=1,566 1,251x1,8890,852x = 1,566 1,25 ,251x 1,609 1,88 ,889x = 1,566 0,638x 638x = 0,043 043 x=0,067 La cantidad de cobre en la aleación es de 0.067g, es decir que la composición porcentual es es de:
%Cu= 0,067 0,852 ×100% %Cu = 7,8%
El valor de y se obtiene por medio de la primera pri mera ecuación:
y=0,852x y=0,8520,067 y=0,785 La composición porcentual del aluminio en la aleación es:
%Al= 0,785 0,852 ×100% %Al = 92,1% SEGUNDO ENSAYO: La aleación de plata y cobre tienen la misma masa que en el primer ensayo, entonces: entonces:
CuAl=0,852→xy=0,852→ y=0,852x1
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
1 mol de Cu × 1 mol de CuO × 79,536g CuO ? gCuO gCuO = xg Cu × 63,546 g Cu 1 mol de Cu 1 mol de CuO = 1,251 1,251gg CuO 3 mol de Al × 1 mol de AlO × 101,959g AlO ?gAlO =yg Al× 126,981g Al 2 mol de Al 1 mol de AlO = 1,88 ,889g AlO 4 Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCuO gCuOx gAlOy=1,567 1,251x1,8890,852x = 1,567 1,25 ,251x 1,609 1,88 ,889x = 1,567 0,638x 638x = 0,042 042 x=0,065 La cantidad de cobre en la muestra es de 0.065g, es decir que la composición porcentual es de:
%Cu= 0,065 0,852 ×100% %Cu = 7,6% El valor de y se obtiene por medio de la primera ecuación:
y=0,852x y=0,8520,065 y=0,787 La composición porcentual del aluminio en la aleación es:
%Al= 0,787 0,852 ×100% %Al = 92,3%
TALLER SKOOG- GRAVIMETRIA Y TITULACIONES DE NEUTRALIZACION. NEUTRALIZACION.
APLICACIONES
DE
LAS
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
tratamiento, el precipitado fue de 0.3181g, calcúlese el porcentaje de cloruro y yoduro en la muestra. Pm yodo (I) = 126.9g/mol Pm Cloro (Cl) = 35.45g/mol g Muestra = 0.6407g Halogenuro de Plata (Ag) = 0.443g Halogenuro de AgCl = 0.3181g
0.443 0.443gg 0.318 0.3181g 1g = 0.124 0.1249g 9g
Estos gramos de diferencia representan la pérdida de gramos de yodo y ganancia de cloro.
126.9g/mol 9g/mol 35.45g/mo 35.45g/moll = 91.45g/mol Se calcula la cantidad de Yodo Yodo (I) que se perdió
0.1249g X 234.8gAgI = 0.320 gAgI= 91.45g/mol 7g AgI AgI 1molAgI 0.3207g 1 mol AgI × 1 mol I × 126.9 g I = 0,17 gI = 0.3207 3207gg AgI AgI × 234.8g 33 AgI 1 mol AgI 1 mol I 0,1733 0.443g 443g 0.32 0.3207 07gg = 0.1223 1223gg Estos gramos de diferencia representan la cantidad de gramos de cloruro de plata.
1 mol AgCl × 1 mol Cl × 35.95gCl = 0.030 gCl gCl = 0.1223 1223gA gAgC gCll × 143.35gAgCl 24gCl gCl 1 mol AgCl 1 mol Cl 0.03024 %Cl= 0.03024gCl X 10 100 4 72%
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
0,1796 g Mg 2 P2 O7 ×
×
0,328 g Mg 2 P2 O7 6,881 g 0,5923 g AgCl × 58,44 g NaCl 1 mol de NaCl 0,242 g NaCl 50 mL 2,42 g NaCl 6,881 g
222,54 g Mg 2 P2 O7
×
2 moles de Mg 1 mol de Mg 2 P2 O7
203,205 g de MgCl 2∙ 6H2 O
0,328 g MgCl 2∙ 6H2 O 50 mL
1 mol de Mg2 P2 O7
1 mol de MgCl 2∙ 6H2 O
×
1 mol de MgCl 2∙ 6H2 O 1 mol de Mg
=0,328 g MgCl 2∙ 6H2 O
×500mL=3,28 g de MgCl 2∙ 6H2 O
×100= 47,666 47,666 % g de MgCl 2∙ 6H2 O 1 mol de AgCl
143,321 g AgCl
×
1 mol de Cl 1 mol de AgCl
×
1 mol de NaCl 1 mol de Cl
=0,242 g NaCl
× 500mL=2,42 g de NaCl
×100= ×100= 35,1 % de NaCl NaCl
6-37. Se analizaron varias aleaciones que solo contienen plata y cobre mediante la disolución de cantidades conocidas en ácido nítrico introduciendo un exceso de IO 3- y llevando la mezcla filtrada de AgIO 3- y Cu(IO 3-)2 a peso constante. Con los datos siguientes calcúlese la composición porcentual de las aleaciones. Masa de la muestra a) 0,2175
Masa del precipitado 0,7391
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
x=Cu y=Ag a) La masa de la muestra contiene 0,2175g de plata y cobre en la aleación:
CuAg=0,2175→xy=0,2175→ y=0,2175x1 → gCuIO−xAgIO−y=0,73912 ?gCu ?gCuIO− =
?gAgIO
−
− 413,3504g Cu − 1 mol Cu 1 mol Cu Cu IO Cu IO = xg Cu × 63,546 g Cu × 1 mol Cu × 1 mol Cu − CuIO − = 6,5047g Cu CuIO 3
− 282,7704 g AgIO − 1 mol de Ag 1 mol de AgIO = yg Ag × 107,8682g Ag × mol de Ag × 1 mol de AgIO − − = 2,6214 6214gg AgI AgIOO 4
Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCu gCuIO−x gAgIO−y=0,7391 6,5047x2,6214 0,2175x = 0,7391 7391 6,5047 5047xx 0,5701 5701 2,6214 6214xx = 0,7391 7391 3,8833 8833xx = 0,1690 1690 x=0,0435 La cantidad de cobre en la aleación es 0,0435g, es decir que la composición porcentual es de:
%Cu= 0,0435 0,2175 ×100% %Cu = 20,00%
El valor de y se obtiene por medio de la primera ecuación:
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
CuAg=0,1948→xy=0,1948→ y=0,1948x1 → gCuIO−xAgIO−y=0,72252 ?gCuIO− =
?gAgIO
−
− 413,3504g Cu IO − 1 mol Cu 1 mol Cu IO = xg Cu × 63,546 g Cu × 1 mol Cu × 1 mol CuIO − − = 6,5047g CuIO 3
− 282,7704 g AgIO − 1 mol de Ag 1 mol de AgIO = yg Ag × 107,8682g Ag × mol de Ag × 1 mol de AgIO − − = 2,6214 6214gg AgI AgIOO 4
Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCuIO−x gAgIO−y=0,7225 6,5047x2,6214 6,5047x2,6214 0,1948x 0,1948x = 0,7225 7225 6,5047 5047xx 0,5106 5106 2,6214 6214xx = 0,7225 7225 3,8833 8833xx = 0,2119 2119 x=0,0545 La cantidad de cobre en la aleación es 0,0545g, es decir que la composición porcentual es de:
%Cu= 0,0545 0,1948 ×100% %Cu = 27,97%
El valor de y se obtiene por medio de la primera ecuación:
y=0,1948x y=0,19480,0545 y=0,1403
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
− 413,3504g Cu − 1 mol Cu 1 mol Cu Cu IO Cu IO = xg Cu × 63,546 g Cu × 1 mol Cu × 1 mol Cu − CuIO = 6,5047g Cu CuIO− 3
?gAgIO
−
− 282,7704 g AgIO − 1 mol de Ag 1 mol de AgIO = yg Ag × 107,8682g Ag × mol de Ag × 1 mol de AgIO − − = 2,6214 6214gg AgI AgIOO 4
Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCu gCuIO−x gAgIO−y=0,7443 6,5047x2,6214 0,2473x = 0,7443 7443 6,5047 5047xx 0,6482 6482 2,6214 6214xx = 0,7443 7443 3,8833 8833xx = 0,0961 0961 x=0,0247 La cantidad de cobre en la aleación es 0,0247g, es decir que la composición porcentual es de:
%Cu= 0,0247 0,2473 ×100% %Cu = 9,98 ,98%
El valor de y se obtiene por medio de la primera ecuación:
y=0,2473x y=0,24730,0247 y=0,2226 La composición porcentual del aluminio en la aleación es:
%Ag= 0,2226 0,2473 ×100%
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
?gAgIO
−
− 282,7704 g AgIO − 1 mol de Ag 1 mol de AgIO = yg Ag × 107,8682g Ag × mol de Ag × 1 mol de AgIO − − = 2,6214 6214gg AgI AgIOO 4
Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCuIO−x gAgIO−y=0,9962 6,5047x2,6214 0,2386x = 0,9962 9962 6,5047 5047xx 0,6254 6254 2,6214 6214xx = 0,9962 9962 3,8833 8833xx = 0,3708 3708 x=0,0954 La cantidad de cobre en la aleación es 0,0954g, es decir que la composición porcentual es de:
%Cu= 0,0954 0,2386 ×100% %Cu = 39,98%
El valor de y se obtiene por medio de la primera ecuación:
y=0,2386x y=0,23860,0954 y=0,1432
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Los valores de obtenidos (3) y (4) se reemplazan en la ecuación (2) al igual que el valor de y en (1), obteniendo:
gCu gCuIO−x gAgIO−y=0,8506 6,5047x2,6214 0,2386x = 0,8506 8506 6,5047 5047xx 0,6254 6254 2,6214 6214xx = 0,8506 8506 3,8833 8833xx = 0,2252 2252 x=0,0579 La cantidad de cobre en la aleación es 0,0579g, es decir que la composición porcentual es de:
%Cu= 0,0579 0,1864 ×100% %Cu = 31,06%
El valor de y se obtiene por medio de la primera ecuación:
y=0,1864x y=0,18640,0579 y=0,1285 La composición porcentual del aluminio en la aleación es:
%Ag= 0,1285 01864 ×100% %A 68,94%
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
EXCESO:
# # = 2,34 34 × 0,0106 010633 # # = 0,024 0,02499
REACCIÓN: Los milimoles equivalentes del hidróxido de sodio son los mismos que de ácido clorhídrico, A partir de esto se puede conocer la cantidad de ácido que reaccionó.
# = 0,6275 6275 0,02 0,0249 49 = 0,6026 6026 = mmol eq + (1) → 2→ 2 SEGUNDA ALICUOTA:
− − → −− −− → ↓ Se hallan los miliequivalentes de NaOH totales, t otales, en exceso. TOTALES:
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
× 1 × 84,0108 ? = 0,1699 × 1 1 1000 1
? = 0,014 0,01433 en 25 mL de muestra. 0,0143 → 25 → 250 250 = 250 250 × 0,0143 25 = 0,143 0,143 ×100 % = 0,143 0,5 % = 28.6 28.6 % Se reemplaza (2) en (1), ( 1), como sabemos cuanta cantidad hay de lo podemos sustituir en (1).
0,6026 = mmol eq + mmol eq 0,602 0,60266 = 0,1699 0,6026 0,6026 0,1699 0,1699 = mmol eq mmol eq = 0,432 0,43277
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
0,5 = = 0,5 = 0,5 0,143 0,229 ? = 0,128 ×100 % = 0,128 0,5 % = 25,6 % Composición porcentual de la muestra:
= 25,6 25,6 % = 28,6 % = 45,8 % 12.42 Calcúlese el volumen de HCl 0.06122 M necesario para titular:
a. 20.00 mL de 0.05555 M al punto final con timolftaleína. timolft aleína. b. 25.00 mL de 0.05555 M al punto final con verde de bromocresol. c. 40.00 mL de una solución que es 0.02102 M en y 0.01655 M en al punto final con verde de bromecresol.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
mmol NaPO = concentra concentración ción NaPO × volume volumenn g NaPO mmol NaPO = 0.05555 M M × 25.00 mL mL mmol NaPO = 1,389 En el punto final, las mmol del son las mismas milimoles del HCl, por lo tanto se puede hallar el volumen necesario de HCl para titular:
mmol de HCl v = concentracion de HCl mmol de HCl = 22,6 v = 1,3890.06122 22,688 mL M Se debe tener en cuenta que en solución el verde bromocresol determina la mitad del componente, por lo tanto, al realizar el cálculo se debe multiplicar por dos para conocer
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
mmol de HCl v = concentracion de HCl 0,8408 × 2 0,662 = 38,2 v = 0.06122 M 0.06122 M 38,288 mL Se debe tener en cuenta que en solución el verde bromocresol determina la mitad del componente por lo tanto, al realizar realizar el cálculo anterior se multiplico por dos para conocer conocer la totalidad del volumen volumen que este requería. requería.
= 38.28 28 d. Timolftaleína.
mmol NaPO = concentra concentración ción NaPO × volume volumenn g NaPO
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
a. 25 ml de una solución que es 0,03000 M en HCl y 0.01000 M en H PO al punto final con verde verde de bromocresol. bromocresol. b. La solución en a) al punto final con timolftaleina. c. 30.00 ml de Na HPO 0.06407 M al punto final final con timolftaleina. d. 25.00 ml de una solución que es 0.02000 M en H PO y 0.03000 M en timolft aleina. NaHPO al punto final con timolftaleina. a. Bromocresol.
mmol mmol HCl = concen concentra tració ciónn HCl × volume volumenn HCl HCl mmol mmol HCl HCl = 0,03000 M × 25.00 mL mL mmol mmol HCl HCl = 0,75 75 luego calculamos las mmol de H PO con el mismo procedimiento anterior anterior para obtener los mmol total total de la mezcla.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
mmol HPO = concen concentra tració ciónn Na NaPO × volume volumenn g NaPO mmol HPO = 0.01000 M × 20.00 mL mmol HPO = 0,25 ,25 En el punto final, las mmol del H PO mas los mmol de HCl son las mismas mmol del NaOH ,por lo tanto tanto se puede hallar hallar el volumen necesario necesario de NaOH para para titular:
mmol de NaOH v = concentracion de NaOH 0,75 0,25 = 12,93 mL v = 0,07731 M 0,07731 M
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
mmol HPO = concen concentra tració ciónn HPO ×volumen HPO mmol HPO = 0.02000 × 25.00 mL mmol HPO = 0,5 En el punto final, las mmol del H PO mas los mmol de Na HPO son las mismas mmol del NaOH, por lo tanto se puede hallar el volumen necesario de NaOH para titular:
mmol de NaOH v = concentracion de NaOH 0,75 0,5 = 16,17 mL v = 0,07731 M 0,07731 M 12.44 Un grupo de soluciones que contiene NaOH,
y , ya sea como
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
a. Al observar la tabla nos percatamos el volumen de acido utilizado con fenolftaleína y verde de bromocresol bromocresol son iguales, iguales, podemos decir gracias gracias a la tabla 16.2 que la solución esta compuesta solo por NaOH,
= = ∗ = 0.1 0.120 2022 ⁄ ∗ 22.42 = 2.69 69
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
=
= .
= 0.07 0.0753 53 ⁄ = . ∗
=. ⁄
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
= 0.1 0.120 2022 ⁄ ∗ 32.23 16.12 = 1.936 1.936 =
= .
= 0.07 0.0774 74 ⁄