Chemistry - Eudiometry Question with Solution | TestHub

The balanced combustion reaction is:

$C_x{H}_{12}(g)+(x+3)O_2(g)\to xC{O}_2(g)+6H_{2}O(g)$

Since the temperature is 400 K, all reactants and products are in the gaseous state.

Let $n_0$ be the initial moles of $C_x{H}_{12}$ and $n_{O_2}$ be the initial moles of $O_2$.

Initial total moles: $n_{initial}=n_0+n_{O_2}$

Final total moles: $n_{final}=x\cdot n_0+6\cdot n_0=(x+6)n_0$ (assuming $n_0$ moles of $C_x{H}_{12}$ reacted completely)

From the balanced equation, for every mole of $C_x{H}_{12}$ reacted, $(x+3)$ moles of $O_2$ are consumed.

Since no reactants are left, the initial moles of $O_2$ must be $n_{O_2}=(x+3)n_0$.

So, $n_{initial}=n_0+(x+3)n_0=(x+4)n_0$.

According to the ideal gas law, $PV=nRT$. Since V and T are constant, $P\propto n$.

Therefore, $\frac{P_{final}}{P_{initial}}=\frac{n_{final}}{n_{initial}}$.

The pressure increased by 20%, so $P_{final}=P_{initial}+0.20P_{initial}=1.20P_{initial}$.

$\frac{1.20P_{initial}}{P_{initial}}=\frac{(x+6)n_0}{(x+4)n_0}$

$1.20=\frac{x+6}{x+4}$

$1.20(x+4)=x+6$

$1.20x+4.8=x+6$

$0.20x=1.2$

$x=\frac{1.2}{0.2}=6$