InterviewSolution
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InterviewSolution
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A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that `16 cm^(3)` of the gas require `48 cm^(3)` of oxygen for complete combustion. This shows that the gas isA. only ethaneB. only etheneC. 1:1 mixture of two gasesD. some unkown mixtures of the two gases |
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Answer» Correct Answer - B `C_(x)H_(y) + (x+(Y)/(4))O_(2)rarr xCO_(2)+(y)/(2)H_(2)O` 1 vol. `(x+(y)/(4))vol` `16 cm^(3) 16 (x+(y)/(4))cm^(3)` Now, we have `16(x+(y)/(4)) = 48` Putting `x = 2` `16(2+(y)/(4))=48` or `y = 4` Hence, the hydrocarbon is `C_(2)H_(4)`. |
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