10 mL of a gaseous hydrocarbon on combustion | vedclass

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(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by:$C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow xCO_2 + \frac{y}{2} H_2O$According to A

Vedclass · Accepted Answer

$14$

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(B) The combustion reaction for a hydrocarbon $C_xH_y$ is given by:$C_xH_y + (x + \frac{y}{4}) O_2 ightarrow xCO_2 + \frac{y}{2} H_2O$According to Avogadro's Law,the volume of gases is proportional to the number of moles.Given: $10 \ mL$ of $C_xH_y$ produces $40 \ mL$ of $CO_2$ and $50 \ mL$ of $H_2O$.From the stoichiometry,$1 \ mol$ of $C_xH_y$ produces $x \ mol$ of $CO_2$ and $\frac{y}{2} \ mol$ of $H_2O$.Thus,$10x = 40 \implies x = 4$.And $10 	imes (\frac{y}{2}) = 50 \implies 5y = 50 \implies y = 10$.The hydrocarbon is $C_4H_{10}$.Total number of atoms $= x + y = 4 + 10 = 14$.

$10 \ mL$ of a gaseous hydrocarbon on combustion gives $40 \ mL$ of $CO_{2(g)}$ and $50 \ mL$ of water vapour. The total number of carbon and hydrogen atoms in the hydrocarbon is:

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