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Estequiometría: Desafío de moles

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Estequiometría: Desafío de moles

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Quiz de estequiometría

Created by

Antonia del Carmen Pérez León
Antonia del Carmen Pérez León
Mexico

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Estequiometría: Desafío de moles
 

Estequiometría: Desafío de molesOnline version

Quiz de estequiometría

by Antonia del Carmen Pérez León
1

En la reacción balanceada 2 H2 + O2 → 2 H2O, ¿cuál es la relación molar entre H2 y H2O?

2

Si se usan 4 mol de N2 y 6 mol de H2 con N2 + 3 H2 → 2 NH3, ¿qué reactivo es limitante y cuántos moles de NH3 se forman?

3

La reacción tiene rendimiento teórico de 10 g y rendimiento real de 7 g. ¿Cuál es el rendimiento porcentual?

4

¿Cuántos moles hay en 24 g de CO2 (MM 44 g/mol)?

5

Una mol de CH4 produce una mol de CO2. ¿Cuánta masa de CO2 se forma a partir 16 g de CH4?

6

Si el rendimiento teórico es 50 g y el rendimiento real es 42 g, ¿Qué porcentaje de rendimiento es?

7

En 2 H2 + N2 → 2 NH3, si tienes 5 g de H2 (MM 2 g) y 28 g de N2 (MM 28 g), ¿qué reactivo es el limitante y cuánto NH3 se forma?

8

En 2 Na + Cl2 → 2 NaCl, si se tienen 3 mol Na y 1 mol Cl2, ¿qué reactivo es el limitante y cuánto de NaCl se forman?

9

En C + O2 → CO2, si tienes 3 mol C y 2 mol O2, ¿cuántos moles de CO2 se pueden formar?

10

¿Cuántos moles hay en 12 g de C (MM 12 g/mol)?

Explicación

La relación es 2:2, es decir 1:1 entre H2 y H2O.

Para 4 N2 se requieren 12 H2, por lo que el H2 es el limitante.

Rendimiento porcentual = (real/theoretical)×100.

moles = masa/MM = 24/44 ≈ 0.545 mol.

1 mol CH4 → 1 mol CO2; 16 g CH4 → 44 g CO2.

Rendimiento = (42/50)×100 = 84%.

H2: 5 g → 2.5 mol; requiere 1.5 mol NH3 por 3 H2; NH3 formado ≈ (2/3)×2.5 ≈ 1.67 mol.

1 Cl2 consume 2 Na; con 1 mol Cl2, se forman 2 mol NaCl por 1 mol Cl2.

O2 es el limitante: 2 mol O2 producen 2 mol CO2.

Moles = masa/MM = 12/12 = 1 mol.

 
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