6 L of an alkene require 27 L of oxygen at | vedclass

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(D) The general combustion reaction for an alkene $(C_nH_{2n})$ is given by:$C_nH_{2n} + \frac{3n}{2} O_2 \rightarrow n CO_2 + n H_2O$According to Avo

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Propene

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(D) The general combustion reaction for an alkene $(C_nH_{2n})$ is given by:$C_nH_{2n} + \frac{3n}{2} O_2 \rightarrow n CO_2 + n H_2O$According to Avogadro's Law,at constant temperature and pressure,the volume ratio is equal to the stoichiometric coefficient ratio.Given: Volume of alkene = $6 \ L$,Volume of $O_2$ = $27 \ L$.Ratio = $\frac{27}{6} = 4.5$.From the equation,the ratio of $O_2$ to alkene is $\frac{3n}{2}$.So,$\frac{3n}{2} = 4.5$.$3n = 9$.$n = 3$.The alkene with $n = 3$ is propene $(C_3H_6)$.

$6 \ L$ of an alkene require $27 \ L$ of oxygen at constant temperature and pressure for complete combustion. The alkene is.....

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