5 , L of an alkane requires 25 , L of oxygen | vedclass

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(D) The general combustion reaction for an alkane $C_nH_{2n+2}$ is:$C_nH_{2n+2} + (\frac{3n+1}{2})O_2 	o nCO_2 + (n+1)H_2O$According to Avogadro's la

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Propane

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(D) The general combustion reaction for an alkane $C_nH_{2n+2}$ is:$C_nH_{2n+2} + (\frac{3n+1}{2})O_2 	o nCO_2 + (n+1)H_2O$According to Avogadro's law,at constant temperature and pressure,the volume of a gas is directly proportional to the number of moles.Therefore,the ratio of volumes is equal to the ratio of stoichiometric coefficients:$\frac{V_{alkane}}{V_{O_2}} = \frac{1}{\frac{3n+1}{2}} = \frac{2}{3n+1}$Given $V_{alkane} = 5 \, L$ and $V_{O_2} = 25 \, L$:$\frac{5}{25} = \frac{2}{3n+1}$$\frac{1}{5} = \frac{2}{3n+1}$$3n+1 = 10$$3n = 9$$n = 3$Since $n=3$,the alkane is propane $(C_3H_8)$.

$5 \, L$ of an alkane requires $25 \, L$ of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure,the alkane is

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