Ejercicios de Estequiometria de las reacciones: Estequiometria de las sustancias y grado de pureza.
Ejercicio N°1. La caliza es una roca sedimentaria compuesta principalmente por carbonato de calcio (CaCO3), generalmente calcita, aunque frecuentemente presenta trazas de magnesita (MgCO3) y otros carbonatos. También puede contener pequeñas cantidades de minerales como arcilla, hematita, siderita, cuarzo, etc., que modifican el color y el grado de coherencia de la roca. El carácter prácticamente monomineral de las calizas permite reconocerlas fácilmente gracias a dos características físicas y químicas fundamentales de la calcita: es menos dura que el cobre (su dureza en la escala de Mohs es de 3) y reacciona con efervescencia en presencia de ácidos tales como el ácido nítrico. En base a lo mencionado anteriormente: Halle la riqueza de CaCO3 que presenta la roca caliza, sabiendo que 0,35 g de esta roca reaccionan con 60 ml de una disolución 0,1 mol/L de ácido nítrico.
Solución.
Datos:
m= 0,35g
V= 60ml ÷ 1000 = 0,06L
mol/l= 0,1mol/L HNO3
La reacción que tiene lugar es:
CaCO3 + 2HNO3 → Ca(NO3)2 + CO2+ H2O
En donde la cantidad de ácido nítrico que se tiene es lo que reaccionará con el carbonato de calcio de la muestra, pero no con el resto de las impurezas, por lo que partiendo de la cantidad de ácido que reacciona, se calcula la cantidad de carbonato de calcio que se tenía en la muestra dada.
La cantidad de ácido que interviene en la reacción se determina a partir de la definición de mol/l de una disolución:
Moles de soluto → 1 litro de solución
0,1 moles → 1 L
X → 0,05 L
X= 0,06 L . 0.1 moles
1 L
X= 0,006moles
Calculamos las masas moleculares de los compuestos involucrados:
Masa molecular CaCO3 = 100g/mol
Masa molecular HNO3 = 63g/mol
De acuerdo con la estequiometria de la reacción, tenemos:
CaCO3+ 2 HNO3 → Ca(NO3)2+ CO2+ H2O
1 mol = 100 g — 2 moles = 2*63 = 126 g
X —— 0,006moles
X= 0,006 moles . 100 g
2 moles
X= 0,30g de CaCO3 que había en la muestra inicial
Como se tenía 0,35g de muestra, la riqueza de la misma es:
% CaCO3 = 0,30g x 100
0,35g
%CaCO3 = 85,71% de riqueza en carbonato de calcio
Ejercicio N° 2. El ácido sulfúrico es el compuesto químico que más se produce en el mundo, por eso se utiliza como uno de los tantos medidores de la capacidad industrial de los países. Una gran parte se emplea en la obtención de fertilizantes. También se usa para la síntesis de otros ácidos como el ácido clorhídrico. El proceso de obtención del ácido clorhídrico a través del ácido sulfúrico consiste fundamentalmente en poner a reaccionar cloruro de sodio con ácido sulfúrico concentrado. Si a través de dicho proceso se obtiene 84g de ácido clorhídrico concentrado con un 33% de pureza. Realice los cálculos estequiométricos para determinar la cantidad de ácido sulfúrico de 72% de pureza que se emplearon en el proceso.
Solución.
Datos:
gramos H2SO4 = ?
% pureza H2SO4= 72%
gramos HCl = 84g
% pureza HCl = 33%
Se plantea y se balancea por tanteo la reacción química:
2 NaCl + H2SO4 → Na2 SO4 + 2 HCl
Luego se determina las masas moleculares de los ácidos sulfúrico y clorhídrico:
Masa molecular H2SO4 = 98g/mol
Masa molecular HCl = 36,5g/mol
Seguidamente se calculan los gramos puros de HCl:
84g ———— 100%
X —————33%
X = 27,72g puros de HCl
Ahora se realiza la relación estequiométrica entre los ácidos sulfúrico y clorhídrico:
H2SO4 ———————- 2HCl
98g/mol 2.(36,5g/mol)
X 27,72g
X = 37,21g H2SO4
Por último, se calcula los gramos puros de H2SO4, haciendo uso de la siguiente relación:
37,21g ————— 100%
X ———————– 72%
X= 26,80 g puros de H2SO4 |
Entonces, son necesarios 26,80 g de H2SO4 para obtener 84g de HCl concentrado de 33% de pureza.
Ejercicio N°3. La actividad industrial produce la emisión de una gran cantidad de gases contaminantes a la atmósfera; vapor de agua, dióxido de carbono, metano, óxidos de nitrógeno, ozono y CFC’s (clorofluorocarburos). Estos gases son los responsables de la calidad del aire que respiramos. Una concentración elevada de gases contaminantes puede producir enfermedades respiratorias e incluso la muerte en los seres vivos. Por tal razón, se ha buscado disminuir la emisión de estos gases en la atmosfera. Una forma efectiva de eliminar los óxidos de nitrógeno de las emisiones gaseosas es hacerlo reaccionar con amoniaco, de acuerdo a la siguiente reacción: NH3 + NO → N2 + H2O. De acuerdo a lo señalado anteriormente: Calcule los gramos de amoníaco que se necesitarán para que reaccionen 12,5 moles de monóxido de nitrógeno.
Solución.
Datos:
gramos NH3 =?
moles NO = 12,5moles
Balanceamos por tanteo la ecuación química proporcionada:
4NH3 + 6NO → 5N2 + 6H2O
La cantidad de amoniaco necesaria para reaccionar con las 12,5 moles de monóxido de nitrógeno se calcula partiendo de la estequiometria de la reacción:
4NH3 + 6NO → 5N2 + 6H2O
4 moles = 4,17 = 68g 6 moles 5 moles 6 moles
X 12,5moles
68g NH3 —————- 6 moles NO
X ————————- 12,5moles NO
X= 141,67gNH3 |
Se necesitan 141,67g de amoniaco para eliminar 12,5 moles de monóxido de nitrógeno.
Ejercicio N°4. Un grupo de estudiantes realizan una práctica de laboratorio con la finalidad de preparar hidróxido de aluminio, el cual es usado para disminuir la acidez estomacal y ayudar a aliviar los síntomas de úlceras, pirosis o dispepsia. Para ello toman una muestra de 15g de sulfato de aluminio y lo hacen reaccionar con 20g de hidróxido de sodio, obteniéndose 3g de hidróxido de aluminio. De acuerdo a lo planteado: calcule el rendimiento porcentual de dicha reacción.
Solución.
Datos:
gramos Al2 (SO4)3 = 15g
gramos NaOH = 20g
gramos Al(OH)3 = 3g
%R=?
Se plantea y balancea por tanteo la ecuación química:
Al2 (SO4)3 (ac) + 6NaOH (ac) → 2 Al(OH)3 (S) + 3 Na2SO4 (ac)
Para realizar los cálculos estequiométricos, calculamos las masas moleculares de los compuestos involucrados:
Masa molecular Al2 (SO4)3 = 342g/mol
Masa molecular NaOH = 40g/mol
Masa molecular Al(OH)3 = 78g/mol
Parte I. Se calcula el reactivo limitante, para ello confronto a los reactantes:
Al2 (SO4)3 ————- 6NaOH
342g/mol 6(40g/mol)
15g X
X = 10,53g NaOH
Al2 (SO4)3 ————- 6NaOH
342g/mol 6(40g/mol)
X 20g
X = 28,50g Al2 (SO4)3
El sulfato de aluminio es el reactivo limitante, porque de 15g que presenta, no puede rendir con 28,50g que necesita.
Parte II. Se calcula el rendimiento teórico de la reacción, para ello se confronta el reactivo limitante con el producto.
Al2 (SO4)3 —————— 2 Al(OH)3
342g/mol 2.(78g/mol)
15g X
X= 6,84g Al(OH)3
Parte III. Se calcula el rendimiento de la reacción haciendo uso de la siguiente fórmula:
%R = Masa experimental x 100
Masa téorica
%R = 3 g x100
6,48 g
%R= 43,86% |
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