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Question

50 mL of a mixture of CO and CH4 was exploded with 85 mL of O2. The volume of CO2 produced was 50 mL. Calculate the percentage composition of the gaseous mixture:

A
CO=20%, CH4=80%
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B
CO=80%, CH4=20%
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C
CO=40%, CH4=60%
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D
CO=60%, CH4=40%
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Solution

The correct option is A CO=20%, CH4=80%
Let, the volume of CO=x mL
Then, volume of CH4=(50x) mL
CO+12O2CO2xmLx2mLxmL ...(i)

CH4+2O2CO2+2H2O(50x)mL2(50x)mL(50x)mL ...(ii)
Total volume of O2=x2+2(50x)=85mL
x=10 mL
Volume of CO = 10 mL
Volume of CH4 = (5010)=40 mL
Percentage of CO = 10×100100=20%
Percentage of CH4 = (10020)=80%

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