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A gas mixture of 3.0 L of propaane and butane on complete combustion at 25^(@)C produced 10 L of CO_(2). Find out the composition of the gas mixture. |
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Answer» Solution :`C_(3)H_(8)+5O_(2)rarr 3CO_(2)+4H_(2)O` `C_(4)H_(10)+6(1)/(2)O_(2)rarr 4CO_(2)+5H_(2)O` Suppose propane = xL. Then butane = (3-x) L `1 L C_(3)H_(8)` PRODUCES 3 L of `CO_(2)` and 1 L of `C_(4)H_(10)` produced 4 L of `CO_(2)` Hence, `CO_(2)` produced `=3x +4(3-x)=12-x=10 L` or x = 2 L `therefore` Propane = 2 L and butane = 3 - 2 = 1 L. |
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