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Standard XII

Chemistry

Stoichiometric Calculations

50 g of ZnS a...

Question

50 g of ZnS are strongly heated in air to effect partial oxidation and the resultant mass weighed $44$ g. What is the ratio of ZnO to ZnS in the resultant mixture.

13.5 : 30.5

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27 : 12.58

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27 : 15.31

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30.52 : 13.48

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Solution

The correct option is D $30.52 : 13.48$
The molar masses of ZnS and ZnO are $97.4$ g/mol and $81.4$ g/mol respectively.
Out of $50$ g of ZnS, let $x$ g of ZnS react.
$50 - x$ g of ZnS remains unreacted.
The mass of ZnO formed $= \frac{81.4 x}{97.4}$ .
The sum of the masses of unreacted ZnS and the product ZnO is $44$ g.
$50 - x + \frac{81.4 x}{97.4} = 44$ $⟹ 50 - x + 0.836 x = 44$
$6 = 0.164 x$
$x = 36.48$ g
Hence, the mass of ZnS that remains unreacted is $50 - x = 50 - 36.48 = 13.52$ g.
The mass of ZnO formed $= 44 - 13.52 = 30.48$ g.
Hence, the ratio of the masses of ZnO to ZnS $= 30.52 : 13.48$ .

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50 g of ZnS are strongly heated in air to effect partial oxidation and the resultant mass weighed 44 g. What is the ratio of ZnO to ZnS in the resultant mixture.

50 g of ZnS are strongly heated in air to effect partial oxidation and the resultant mass weighed $44$ g. What is the ratio of ZnO to ZnS in the resultant mixture.