0.5 L of gaseous hydrocarbon when burnt in e | vedclass

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(C) According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to the number of moles $(V \propto n

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$C_7H_{16}$

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(C) According to Avogadro's Law,at constant temperature and pressure,the volume of gases is directly proportional to the number of moles $(V \propto n)$.Let the hydrocarbon be $C_xH_y$.The combustion reaction is: $C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O$.Given volumes:$V_{hydrocarbon} = 0.5 \ L$$V_{CO_2} = 3.5 \ L$$V_{H_2O} = 4 \ L$From the stoichiometry:$x = V_{CO_2} / V_{hydrocarbon} = 3.5 / 0.5 = 7$$y/2 = V_{H_2O} / V_{hydrocarbon} = 4 / 0.5 = 8 \Rightarrow y = 16$Thus,the molecular formula is $C_7H_{16}$.

$0.5 \ L$ of gaseous hydrocarbon when burnt in excess of $O_2$ gave $3.5 \ L$ of $CO_2$ and $4 \ L$ of water vapours under same conditions. The molecular formula of the hydrocarbon will be:

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