Determine the volume of gas released when dissolving 50 g of iron sulfide (2) containing 12%

| August 15, 2021 | education

Determine the volume of gas released when dissolving 50 g of iron sulfide (2) containing 12% impurities in an excess of sulfuric acid.

Let’s find the mass of pure iron sulfide without impurities.

100% – 12% = 88%.

50g – 100%,

X – 88%,

X = (50 g × 88%): 100 = 44 g.

Find the amount of iron sulfide substance by the formula:

n = m: M.

M (FeS) = 56 + 32 = 88 g / mol.

n = 44 g: 88 g / mol = 0.5 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

FeS + H2SO4 = FeSO4 + H2S ↑.

According to the reaction equation, 1 mole of iron sulfide accounts for 1 mole of hydrogen sulfide. The substances are in quantitative ratios of 1: 1. The amount of the substance of iron sulfide and hydrogen sulfide will be the same.

n (FeS) = n (H2S) = 0.5 mol.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.5 mol × 22.4 L / mol = 11.2 L.

Answer: 11.2 liters.



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