LAQ : What amount of ferrous sulphide will be required to prepare 1.7 gm of H2S by reaction of Ferrous sulphide with excess dilute H2SO4?

Question

What amount of ferrous sulphide will be required to prepare 1.7 gm of H₂S by reaction of Ferrous sulphide with excess dilute H2SO₄? [Fe = 56, S = 32, H = 1, 0 = 16]

Answer

Balanced chemical equation:

FeS + dil H₂SO₄ = FeSO₄ + H₂S

Molecular Mass of FeS = 56 + 32 = 88 g

Molecular Mass of H₂S = 1×2+32 = 34 g

It can be seen from the equation that 34 gm of H₂S is produced from 88 gm of FeS

∴ To prepare 1.7 gm of H₂S, the amount of FeS needed

= 88×1.7\over 34

= 4.4 gm