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`10mL` of gaseous hydrocarbon on combustion gives `40ml` of `CO_(2)(g)` and `50mL` of `H_(2)O` (vapour) The hydrocarbon is .A. `C_(4)H_(5)`B. `C_(8)H_(10)`C. `C_(4)H_(8)`D. `C_(4)H_(10)` |
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Answer» `{:(C_(a)H_(b),+,(a+(b)/(4))O_(2),rarr,aCO_(2),+,(b)/(2)H_(2)O,),(10,rarr,"excess",,-,,-,),(0,,"excess",,10a,,5b,):}` Therefore, `10 a = 40` So, `a = 4 " " rArr 5b = 50 rArr b = 10` `:.` (D) |
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