# Calculate the volume under normal conditions of a gas mixture consisting of 0.3 mol H2

Calculate the volume under normal conditions of a gas mixture consisting of 0.3 mol H2, 0.4 mol N2 and 0.3 mol CO.

Given:
n (H2) = 0.3 mol
n (N2) = 0.4 mol
n (CO) = 0.3 mol

Find:
V (mixture) -?

1) Calculate the volume of H2 (under normal conditions):
V (H2) = n (H2) * Vm = 0.3 * 22.4 = 6.72 liters;
2) Calculate the volume N2 (under normal conditions):
V (N2) = n (N2) * Vm = 0.4 * 22.4 = 8.96 L;
3) Calculate the volume of CO (under normal conditions):
V (CO) = n (CO) * Vm = 0.3 * 22.4 = 6.72 liters;
4) Calculate the volume of the mixture (under normal conditions):
V (mixture) = V (H2) + V (N2) + V (CO) = 6.72 + 8.96 + 6.72 = 22.4 liters.

Answer: The volume of the mixture is 22.4 liters.

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