Zinc blende [ZnS] is roasted in air. Calculate :

(a) The number of moles of sulphur dioxide liberated by 776 g of ZnS and

(b) The weight of ZnS required to produce 22.4 lits. of SO₂ at s.t.p. [S = 32, Zn = 65, O = 16]

$\underset{65 + 32 = 97 \text{ g}}{\text{ZnS}} + \text{O}$2 \longrightarrow \underset{1 \text{ mole}}{\text{SO}2}

97 g of ZnS gives 1 mole of SO₂

∴ 776 g of ZnS will give $\dfrac{1}{97}$ x 776 = 8 moles of SO₂

Hence, 8 moles of sulphur dioxide is liberated by 776 g of ZnS

(ii)

$\underset{65 + 32 = 97 \text{ g}}{\text{ZnS}} + \text{O}$2 \longrightarrow \underset{1 \text{ mole}}{\text{SO}2}

97 g of ZnS liberates 1 mole of SO₂

As 1 mole occupies 22.4 lit volume at s.t.p.

∴ 22.4 lit of SO₂ is produced by 97 g of ZnS.