(a) ZnS + O2 → SO2
97 g 1 mole
97 g of ZnS gives 1 mole of SO2
∴ 776 g of ZnS will give 1\over 97 × 776 = 8 moles of SO2
Hence, 8 moles of sulphur dioxide is liberated by 776 g of ZnS
(b) ZnS + O2 → SO2
97 g 1 mole
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.