Question

How much amount of ferrous sulphide reacts with dilute H,SO, to get 1.12 L H,S at NTP?​

Answer 1

Question : How much amount of ferrous sulphide reacts with dilute H₂SO₄ to get 1.12 L H₂S at NTP ?

solution : volume of H₂S = 1.12 L

no of moles of H₂S at NTP = given volume/22.4 L

= 1.12/22.4 L

= 1/20 = 0.05 mol

reaction of ferrous sulphide , (FeS) with dilute H₂SO₄ gives ferrous sulphate, FeSO₄ and hydrogen sulphde, H₂S.

i.e., FeS + H₂SO₄ => FeSO₄ + H₂S

it is clear that 1 mole of FeS is needed to get 1 mole of H₂S.

so, 0.05 mole of FeS will be needed to get 0.05 mole of H₂S.

now mass of FeS = mole of FeS × molar mass of FeS

= 0.05 mol × 88 g/mol

= 4.4 g

Therefore 4.4g of ferrous sulphide reacts with dilute H₂SO₄ to get 1.12 L H₂S at NTP.