(a) ZnS + O_{2} → SO_{2}

97 g 1 mole

97 g of ZnS gives 1 mole of SO_{2}

∴ 776 g of ZnS will give 1\over 97 × 776 = 8 moles of SO_{2}

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

(b) ZnS + O_{2} → SO_{2}

97 g 1 mole

97 g of ZnS liberates 1 mole of SO_{2}

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

∴ **22.4 lit of SO _{2} is produced by 97 g of ZnS.**