Question

calculate the amount of hydrogen sulphide that would be liberated by the action of sulphuric acid on 4.4 gram of ferrous sulphide. (Fe=56;S=32)​

Answer 1

You will get iron sulfate and hydrogen sulfide gas (a.k.a. rotten egg gas).

FeS + H2SO4 = FeSO4 + H2S(g)

Change in Free Energy: ΔG(20C) = -66.6kJ (negative, so the reaction runs)

Change in Enthalpy: ΔH(20C) = -37.3kJ (negative, so the reaction is exothermic)

Answer 2

1.7 g of Hydrogen sulphide (H₂S) will be obtained.

We'll begin by writing the balanced equation for the reaction between ferrous sulphide (FeS) and sulphuric acid (H₂SO₄). This is illustrated below:

FeS + H₂SO₄ —> H₂S + FeSO₄

Next, we shall determine the mass of FeS that reacted and the mass of H₂S produced from the balanced equation. This is illustrated below:

Molar mass of FeS = 56 + 32

= 88 g/mol

Mass of FeS from the balanced equation = 1 × 88 = 88g

Molar mass of H₂S = (2×1) + 32

= 2 + 32

= 34 g/mol

Mass of H₂S from the balanced equation = 1 × 34 = 34 g

SUMMARY

From the balanced equation above,

88 g of FeS reacted to produce 34 g of H₂S.

Finally, we shall determine the mass of H₂S produced by the reaction of 4.4 g of FeS. This can be obtained as follow:

From the balanced equation above,

88 g of FeS reacted to produce 34 g of H₂S.

Therefore, 4.4 g of FeS will react to produce = (4.4 × 34)/88 = 1.7 g of H₂S.

Thus, 1.7 g of H₂S were obtained from the reaction.

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