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(C) The combustion reaction for a hydrocarbon $C_{x}H_{y}$ is given by:$C_{x}H_{y(g)} + (x + \frac{y}{4}) O_{2(g)} \rightarrow x CO_{2(g)} + \frac{y}{

Vedclass · Accepted Answer

$8$

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(C) The combustion reaction for a hydrocarbon $C_{x}H_{y}$ is given by:$C_{x}H_{y(g)} + (x + \frac{y}{4}) O_{2(g)} \rightarrow x CO_{2(g)} + \frac{y}{2} H_{2}O(\ell)$According to Avogadro's Law,the volume ratio is equal to the stoichiometric coefficient ratio.Given that $1$ volume of $C_{x}H_{y}$ produces $4$ volumes of $CO_{2}$,we have $x = 4$.Given that $1$ volume of $C_{x}H_{y}$ requires $6$ volumes of $O_{2}$,we have $(x + \frac{y}{4}) = 6$.Substituting $x = 4$ into the equation:$4 + \frac{y}{4} = 6$$\frac{y}{4} = 2$$y = 8$.

The formula of a gaseous hydrocarbon which requires $6$ times of its own volume of $O_{2}$ for complete oxidation and produces $4$ times its own volume of $CO_{2}$ is $C_{x}H_{y}$. The value of $y$ is $......$ .

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