How much hydrogen is required to interact with iron (III) oxide weighing 6.4 g?

Given:

m (Fe2O3) = 6.4 g

Find:

V (H2) -?

Solution:

1) We compose the reaction equation corresponding to the condition of the problem:

Fe2O3 + 3H2 = 3H2O + 2Fe;

2) Find the amount of iron trioxide, 6.4 grams of oxide:

n (Fe2O3) = m: M = 6.4 g: 160 g / mol = 0.04 mol;

3) We compose a logical expression:

if 1 mol of Fe2O3 requires 3 mol of H2,

then 0.04 mol Fe2O3 will require x mol H2,

then x = 0.12 mol.

4) Find the volume of hydrogen required to reduce iron trioxide to metal:

V (H2) = n * Vm = 0.12 mol * 22.4 L / mol = 2.688 L;

Answer: V (H2) = 2.688 l.