InterviewSolution
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InterviewSolution
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At `300 K` and `1 atm, 15 mL` of a gaseous hydrocarbon requires `375 mL` air containing `20% O_(2)` by volume for complete combustion. After combustion, the gases occupy `330 mL`. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon isA. `C_(3)H_(8)`B. `C_(4)H_(8)`C. `C_(4)H_(10)`D. `C_(3)H_(6)` |
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Answer» `C_(x)H_(y)(g)+underset(75 mL)((x+(y)/(4)))O_(2)(g)rarrxCO_(2)underset(30 mL)((g))+(y)/(2)H_(2)O(l)` `O_(2)` used `= 20%` of `375 = 75 mL` Inert part of air `= 80%` of `375 = 300 mL` Total volume of gases `= CO_(2)`+ Inert part of air `= 30 + 300 = 330 mL` `(x)/(1) = (30)/(15) rArr x = 2` `(x+y)/(1) = (75)/(15) rArr x+(y)/(4) = 5` `rArr x = 2, y = 12 rArr C_(2)H_(12)` |
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