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Question

Hg and I2 are allowed to react to form Hg2I2(assume no other side products to be formed). 50 g of Hg is exposed to excess I2 and 65 g of Hg2I2 is obtained. Calculate the percentage yield of the reaction.
(Molar mass of Hg and I are 200 g/mol and 127 g/mol respectively)
2Hg+I2Hg2I2

A
20.5 %
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B
79.5 %
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C
40.8 %
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D
59.6 %
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Solution

The correct option is B 79.5 %
2Hg+I2Hg2I2
Number of mole of Hg = Given massMolar mass=50200=0.25 mol
1 mol of Hg2I2 is produced from 2 mol of Hg
Since, I2 is taken in excess
Theoretically, 0.25 mol of Hg will produce 0.125 mol of Hg2I2
Theoretical yield = number of moles of Hg2I2× molar mass of Hg2I2
Theoretical yield =0.125×654=81.75 g
Now, we know that
% yield=Actual yieldTheoretical yield×100=6581.75=79.5%
Therefore, the percentage yield of the given reaction is 79.5 %

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