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V125BC [204]
3 years ago
12

Un recipiente cerrado, de 4,25 L, con tapa móvil, contiene H2S(g) a 740 Torr y 50,0°C. Se introduce en ese recipiente N2(g) a te

mperatura y presión constantes, de manera que el volumen final es el doble del volumen inicial. Calcular la cantidad de N2(g) en el recipiente, expresada en moles.
porfi ayuda
Chemistry
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

n_{N_2}=6.41mol

Explanation:

¡Hola!

En este caso, teniendo en cuenta la información dada por el problema, inferimos que primero se debe usar la ecuación del gas ideal con el fin de calcular las moles de gas que se encuentran al inicio del experimento:

PV=nRT\\\\n=\frac{RT}{PV} \\\\n=\frac{0.08206\frac{atm*L}{mol*K}*(50.0+273.15)K}{740/760atm*4.25L}\\\\n=6.41mol

Seguidamente, usamos la ley de Avogadro para calcular las moles finales, teniendo el cuenta que el volumen final es el doble del inicial (8.50 L):

n_2=\frac{6.41mol*8.50L}{4.25L}\\\\n_2=12.82mol

Quiere decir que las moles de N2(g) que se agregaron son:

n_{N_2}=12.81mol-6.41mol\\\\n_{N_2}=6.41mol

¡Saludos!

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