Calculate the volume of gas that will be released when an excess of iron (II) sulfide is applied to 500 g of a 12% sulfuric acid solution.

1) Let’s compose the reaction equation:
FeS + H2SO4 = FeSO4 + H2S ↑
2) Calculate the mass and amount of the sulfuric acid substance contained in the solution:
m (H2SO4) = (m (solution) ∙ ω) / 100% = (500 ∙ 12) / 100 = 60 g;
n (H2SO4) = m (H2SO4) / M (H2SO4) = 60/98 = 0.61 mol.
3) Determine the volume of hydrogen sulfide:
according to the reaction equation
n (H2S) = n (H2SO4) = 0.61 mol;
V (H2S) = n (H2S) ∙ Vm = 0.61 ∙ 22.4 = 13.66 l.
Answer: 13.66 liters.

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