Chemistry - Eudiometry Question with Solution | TestHub

Question Explanation: Finding the hydrocarbon class with a specific ${\mathrm{O}}_2/{\mathrm{CO}}_2$ ratio.

Concept: Stoichiometry of Hydrocarbon Combustion.

Solution:

Write the general balanced combustion equation for each class of hydrocarbon (alkane, alkene, alkyne). Then, find the ratio of the volume of oxygen required to the volume of carbon dioxide produced and see which class matches the given ratio of 1.5.

Given: The ratio of $\frac{V_{{\mathrm{O}}_2}required}{V_{{\mathrm{CO}}_2}produced}=1.5=\frac{3}{2}$.

For Alkanes (${\mathrm{C}}_n{\mathrm{H}}_{2n+2}$):

${\mathrm{C}}_n{\mathrm{H}}_{2n+2}+\left(\frac{3n+1}{2}\right){\mathrm{O}}_2⟶n{\mathrm{CO}}_2+(n+1){\mathrm{H}}_2\mathrm{O}$

The ratio is $\frac{(3n+1)/2}{n}=\frac{3n+1}{2n}$.

Setting this equal to the given ratio:

$\frac{3n+1}{2n}=\frac{3}{2}$

$2(3n+1)=3(2n)$

$6n+2=6n$

$2=0$ (Impossible).

For Alkenes (${\mathrm{C}}_n{\mathrm{H}}_{2n}$):

${\mathrm{C}}_n{\mathrm{H}}_{2n}+\left(\frac{3n}{2}\right){\mathrm{O}}_2⟶n{\mathrm{CO}}_2+n{\mathrm{H}}_2\mathrm{O}$

The ratio is $\frac{3n/2}{n}=\frac{3}{2}$.

This ratio matches the given ratio of $\frac{3}{2}$ for any value of $n$.

For Alkynes (${\mathrm{C}}_n{\mathrm{H}}_{2n-2}$):

${\mathrm{C}}_n{\mathrm{H}}_{2n-2}+\left(\frac{3n-1}{2}\right){\mathrm{O}}_2⟶n{\mathrm{CO}}_2+(n-1){\mathrm{H}}_2\mathrm{O}$

The ratio is $\frac{(3n-1)/2}{n}=\frac{3n-1}{2n}$.

Setting this equal to the given ratio:

$\frac{3n-1}{2n}=\frac{3}{2}$

$2(3n-1)=3(2n)$

$6n-2=6n$

$-2=0$ (Impossible).

Therefore, the hydrocarbon class that matches the given ratio is Alkene.

Final Answer:

The correct option is (B)