Calculate the volume of hydrogen reacted with iron (III) oxide if 250 g of iron was formed.

The reduction reaction of iron (III) oxide is described by the following chemical equation:

2Fe2O3 + 6H2 = 4Fe + 6H2O;

To obtain 4 moles of iron, 6 moles of hydrogen are needed.

Let’s find the amount of substance in 250 grams of iron.

Its molar mass is 56 grams / mol;

N Fe = 250/56 = 4.4643 mol;

The required amount of hydrogen will be:

N H2 = 4.4643 x 6/4 = 6.6965 mol;

Under normal conditions, one mole of ideal gas takes up a volume of 22.4 liters.

6.6965 mol of hydrogen will occupy the volume:

V H2 = 6.6965 x 22.4 = 150 liters;