How many liters of hydrogen will it take to reduce 57.6 grams of copper oxide (1) if the reaction produces copper and water?

m = 57.6 g Chl-?

Cu2O + H2 = 2Cu + H2O – OBP, copper and water were formed;
Let’s make the calculations:
M (Cu2O) = 143 g / mol;

M (H2) = 2 g / mol.

Determine the amount of oxide:
Y (Cu2O) = m / M = 57.6 / 143 = 0.4 mol;

Y (H2) = 0.4 mol since the amount of substances is 1 mol.

We find the volume of H2:
V (H2) = 0.4 * 22.4 = 8.96 L

Answer: to reduce copper oxide (1), hydrogen will be required with a volume of 8.96 liters