What mass of iron sulfide 2, containing 12% impurities, is required to obtain 2.8 liters of hydrogen sulfide?

To solve the problem, write down the data:

X g -? 12% impurity.

FeS + 2HCl = H2S + FeCl2 – ion exchange, hydrogen sulfide is released;
V = 2.8 l

Let’s make the calculations:
M (FeS) = 87.8 g / mol;

M (H2S) = 34 g / mol.

Proportion:
1 mole of gas at normal level – 22.4 liters;

X mol (H2S) -2.8 L hence, X mol (H2S) = 1 * 2.8 / 22.4 = 0.125 mol.

We find the mass of the original substance:
m (FeS) = 0.125 * 87.8 = 10.98 g (with impurities);

m (FeS) = 10.98 * (1 – 0.12) = 9.7 g (without impurity).

Answer: you need iron sulfide weighing 9.7 g