ZnS is roasted in air, calculate:

a] the number of moles of SO2 liberated by 776g 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)

Dear Student,

Roasting of ZnS can be represented as

2ZnS + 3O2 → 2ZnO + 2SO2

Molar mass of ZnS = 97.4 g/mol

Molar mass of SO2 = 64 g/mol

Now, according to balanced chemical equation,

(2x97.4) = 194.8 g of ZnS gives (2x64) = 128 g of SO2

So, 776 g of ZnS will give (128x776)/194.8 = 509.9 g of SO2

64 g of SO2 = 1 mole 

509.9 g of SO2 = (509.9/64) = 7.97 mole of SO2.

b)  1 Mole of SOcontains 22.4L of volume irrespective of it's molecular mass. Therefore,  2 moles of SO will have

1 mole SO2 = 22.4L of gas

2 moles of SO2= 2 x 22.4L =44.8L of SO2
From the above roasting equation 2 moles of ZnS produces 2 moles of SO. Therefore,weight of ZnS required to produce 22.4L of SO2 at STP will be,

2 x molecular mass of Zns 44.8LX22.4L=2 x 97.4g 44.8LX22.4L=97.4g


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