A mixture ofCH4 and C2H4 was completely burnt in excess of oxygen, yielding equal volumes of CO2 and steam. Calculate the percentages of the compounds in the original mixture:
A
25% CH4 and 75% C2H4
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B
30% CH4 and 70% C2H4
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C
75% CH4 and 25% C2H4
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D
53.33% CH4 and 46.67% C2H4
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Solution
The correct option is D 53.33% CH4 and 46.67% C2H4 we have,
CH4+O2→CO2+2H2O
and, C2H4+O2→2CO2+2H2O
if they are producing equal volume of CO2 gas, that means moles of CH4 is double the moles of C2H4
let, moles of CH4 = 1, and mass becomes = 16 g
and moles of C2H4 = 0.5 and mass becomes = 14 g
so, percentage composition of C2H4 = 1430∗100 = 46.6 %