Chemistry - Eudiometry Question with Solution | TestHub

Let the combustion reaction be:

$C_x{H}_y(g)+(x+\frac{y}{4})O_2(g)\to xC{O}_2(g)+\frac{y}{2}{H}_{2}O(l)$

Given volumes:

Volume of hydrocarbon ($V_{C_x{H}_y}$) = 15 mL

Volume of $O_2$ initially ($V_{O_2,initial}$) = 100 mL

Total residual gas volume ($V_{total,residual}$) = 77.5 mL

Volume absorbed by KOH ($V_{KOH}$) = 45 mL

KOH absorbs $C{O}_2$. So, $V_{C{O}_2}=45mL$.

From the stoichiometry, 1 volume of $C_x{H}_y$ produces $x$ volumes of $C{O}_2$.

$x=\frac{V_{C{O}_2}}{V_{C_x{H}_y}}=\frac{45mL}{15mL}=3$.

The residual gas after KOH treatment is unreacted $O_2$ ($V_{O_2,unreacted}$) and any other inert gases (none here).

$V_{O_2,unreacted}=V_{total,residual}-V_{C{O}_2}=77.5mL-45mL=32.5mL$.

Volume of $O_2$ reacted ($V_{O_2,reacted}$) = $V_{O_2,initial}-V_{O_2,unreacted}=100mL-32.5mL=67.5mL$.

From the stoichiometry, 1 volume of $C_x{H}_y$ reacts with $(x+\frac{y}{4})$ volumes of $O_2$.

$x+\frac{y}{4}=\frac{V_{O_2,reacted}}{V_{C_x{H}_y}}=\frac{67.5mL}{15mL}=4.5$.

Substitute $x=3$:

$3+\frac{y}{4}=4.5$

$\frac{y}{4}=1.5$

$y=6$.

Therefore, the hydrocarbon is $C_3{H}_6$.

The value of $(x+y)=3+6=9$.