One mole of an alkane (x) requires 8 mole oxy | vedclass

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(B) $1$. The general combustion reaction for an alkane $C_nH_{2n+2}$ is: $C_nH_{2n+2} + (\frac{3n+1}{2})O_2 \rightarrow nCO_2 + (n+1)H_2O$.$2$. Accord

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$17$

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(B) $1$. The general combustion reaction for an alkane $C_nH_{2n+2}$ is: $C_nH_{2n+2} + (\frac{3n+1}{2})O_2 \rightarrow nCO_2 + (n+1)H_2O$.$2$. According to the problem, the coefficient of $O_2$ is $8$, so: $\frac{3n+1}{2} = 8$.$3$. Solving for $n$: $3n + 1 = 16 \Rightarrow 3n = 15 \Rightarrow n = 5$.$4$. The alkane is $C_5H_{12}$ (pentane).$5$. The sum of carbon and hydrogen atoms is $n + (2n + 2) = 5 + 12 = 17$.

One mole of an alkane $(x)$ requires $8$ mole oxygen for complete combustion. The sum of the number of carbon and hydrogen atoms in the alkane $(x)$ is . . . . . . .

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