What is the volume of hydrogen under normal conditions? necessary for the reaction of interaction with 32 g of iron oxide 3?

1.Let’s find the amount of iron (III) oxide substance – Fe2O3.

n = m: M.

M (Fe2O3) = 56 × 2 + 16 × 3 = 160 g / mol.

n = 32 g: 160 g / mol = 0.2 mol.

2. Let’s compose the reaction equation, find the quantitative relations of substances.

Fe2O3 +3 H2 = 2Fe + 3H2O.

For 1 mole of iron oxide, there are 3 moles of hydrogen. The substances are in quantitative ratios of 1: 3. The amount of hydrogen will be 3 times more than the amount of iron oxide.

n (Fe2O3) = 3n (H2) = 0.2 × 3 = 0.6 mol.

3.Let’s find the volume of hydrogen.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.6 mol × 22.4 L / mol = 13.44 L.

Answer: 13.44 liters.