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Question

58 g of butane on combustion gave 158.4 g of CO2. The combustion of butane follows the given reaction:
2C4H10+13O28CO2+10H2O

Calculate the yield of the combustion reaction.

A
50%
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B
70%
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C
80%
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D
90%
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Solution

The correct option is D 90%
2C4H10+13O28CO2+10H2O
Number of moles of butane =5858=1 mol
As per stoichiometry,
2 mol of butane produces 8 mol of CO2
1 mol of butane will produce 82=4 mol of CO2
So, theoretical yield = 4 × molar mass of CO2
theoretical yield =4×44=176 g
% yield=Experimental yieldTheoretical yield×100
% yield=158.4176×100=90%

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