Combustion of 10 mL of a gaseous hydrocarbon | vedclass

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(D) According to Avogadro's Law,the volume ratio of gases is equal to the mole ratio under the same conditions of temperature and pressure.Let the hyd

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$C_4H_{10}$

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(D) According to Avogadro's Law,the volume ratio of gases is equal to the mole ratio under the same conditions of temperature and pressure.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 volume ratio: $V(C_xH_y) : V(CO_2) : V(H_2O) = 10 : 40 : 50 = 1 : 4 : 5$.From the stoichiometry,$x = 4$ (moles of $CO_2$ per mole of hydrocarbon).Also,$y/2 = 5$,which implies $y = 10$.Therefore,the molecular formula is $C_4H_{10}$.

Combustion of $10 \ mL$ of a gaseous hydrocarbon gives $40 \ mL$ of $CO_2$ and $50 \ mL$ of water vapour under the same conditions. The molecular formula of the hydrocarbon is

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