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50 " mL of " a gaseous hydrocarbon A requried for complete combustion. 357 " mL of " air `(21%` oxygen by volume) and gaseous products occupied 327 mL (all volumes being measured at STP. The molecular formula of the hydrocarbon A is (a). `C_2H_6` (b). `C_2H_4` (c). `C_3H_6` (d). `C_3H_6` |
Answer» Let the formula of the hydrocarbon A be CxHy. `C_(x)H_(y)(g)+(x+(y)/(4))O_2(g)toxCO_2(g)+(y)/(2)H_2O(1)` Volume of `O_2=(357xx21)j/(j10j0)=75mL` Volume of `N_2=357-75=282ml` volume of `CO_2=327-282=45mL` `therefore15x=45`, `x=3` `({:(15(x+(y)/(4))=75),(x+(y)/(4)=5):})` `3+(y)/(4)=5` On solving we get `y=8` Formula of A is `C_3H_8` |
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