How much hydrogen is needed to reduce manganese from 78g of manganese (IV) oxide containing 9% impurities?

1. Let’s compose a chemical equation, not forgetting to put down the coefficients:

MnO2 + 2H2 = Mn + 2H2O.

2. Let’s find the mass fraction of manganese oxide:

ω = 100% – 9% = 91% (0.91).

3. Find the mass of manganese oxide:

m (MnO2) = 78g * 0.91 = 70.98g.

4. Find the amount of pure manganese oxide substance:

n (MnO2) = m / M = 70.98g / 87 g / mol = 0.816 mol.

5. The amount of hydrogen is denoted by X. Let’s make the equation:

X = 0.816 mol * 2 mol / 1 mol = 1.632 mol.

6. Find the volume of hydrogen:

V (H2) = 1.632 mol * 22.4 L / mol = 36.55 L.

Answer: V (H2) = 36.55l.