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Question

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]

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Answer

(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.

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