Sodium sulfide weighing 8 g, containing 2.5% impurities, was acted upon by an excess of sulfuric acid solution.

Sodium sulfide weighing 8 g, containing 2.5% impurities, was acted upon by an excess of sulfuric acid solution. At the same time, gas with an unpleasant odor was released. Calculate the volume of gas produced.

1. Let’s compose the reaction equation:

Na2S + H2SO4 = Na2SO4 + H2S ↑;

2.find the mass of pure sodium sulfide:

m (Na2S) = w (Na2S) * m (sample);

m (Na2S) = (1 – 0.025) * 8 = 0.975 * 8 = 7.8 g;

3.Calculate the chemical amount of sulfide:

n (Na2S) = m (Na2S): M (Na2S);

M (Na2S) = 2 * 23 + 32 = 78 g / mol;

n (Na2S) = 7.8: 78 = 0.1 mol;

4.Calculate the amount and volume of released hydrogen sulfide:

n (H2S) = n (Na2S) = 0.1 mol;

V (H2S) = n (H2S) * Vm = 0.1 * 22.4 = 2.24 dm3.

Answer: 2.24 dm3.