10 mL of a gaseous hydrocarbon on combustion | vedclass

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(D) The combustion reaction for a hydrocarbon $C_xH_y$ is: $C_xH_y(g) + (x + \frac{y}{4}) O_2(g) \rightarrow x CO_2(g) + \frac{y}{2} H_2O(g)$ Given vo

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$C_4H_{10}$

Vedclass · Answer

(D) The combustion reaction for a hydrocarbon $C_xH_y$ is: $C_xH_y(g) + (x + \frac{y}{4}) O_2(g) \rightarrow x CO_2(g) + \frac{y}{2} H_2O(g)$ Given volumes: $10 \ mL$ of hydrocarbon produces $40 \ mL$ of $CO_2$ and $50 \ mL$ of $H_2O$. According to Avogadro's Law,volume is proportional to the number of moles at constant temperature and pressure. Ratio of volumes: $1 : x : \frac{y}{2} = 10 : 40 : 50$ Dividing by $10$: $1 : 4 : 5$ Comparing the coefficients: $x = 4$ $\frac{y}{2} = 5 \Rightarrow y = 10$ Thus,the hydrocarbon is $C_4H_{10}$.

$10 \ mL$ of a gaseous hydrocarbon on combustion gives $40 \ mL$ of $CO_{2(g)}$ and $50 \ mL$ of $H_2O_{(g)}$. The hydrocarbon is $-$

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