Chemistry - Stoichiometry Question with Solution | TestHub

Question Explanation: Deducing alkane formula from combustion volume ratios.

Concept: Stoichiometry of Alkane Combustion.

Solution: Write the general balanced equation for alkane combustion. Find the theoretical ratio of the volume of ${\mathrm{O}}_2$ required to the volume of ${\mathrm{CO}}_2$ produced in terms of 'n'. Set this expression equal to the given ratio (7/4) and solve for 'n'.

Let the molecular formula of alkane be ${\mathrm{C}}_n{\mathrm{H}}_{2n+2}$.

The balanced combustion equation is:

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

If 'v' is the volume of the alkane, then the volumes are:

v $\left(\frac{3n+1}{2}\right)vnv\times$

Given:

$$\frac{volof{\mathrm{O}}_2}{vol.of{\mathrm{CO}}_2}=\frac{7}{4}$$

Substitute the theoretical volumes:

$$\frac{\left(\frac{3n+1}{2}\right)v}{nv}=\frac{7}{4}$$

Simplify the equation:

$$4\left(\frac{3n+1}{2}\right)=7n$$

$$2(3n+1)=7n$$

$$6n+2=7n$$

$$n=2$$

So, the molecular formula of the alkane is ${\mathrm{C}}_2{\mathrm{H}}_6$.

Final Answer: ${\mathrm{C}}_2{\mathrm{H}}_6$