Chemistry - Stoichiometry Question with Solution | TestHub

Volume of ${\mathrm{CO}}_2$ produced $=80\mathrm{mL}$

Volume of ${\mathrm{N}}_2$ produced $=60\mathrm{mL}$

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

${\mathrm{N}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}+\left(\frac{\mathrm{x}}{2}\right){\mathrm{O}}_2⟶\frac{\mathrm{x}}{2}{\mathrm{N}}_2+(\mathrm{x}){\mathrm{H}}_2\mathrm{O}$

Volume of ${\mathrm{C}}_{\mathrm{y}}{\mathrm{H}}_{2\mathrm{y}+2}=\frac{80}{\mathrm{y}}$ and Volume of ${\mathrm{N}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}=\frac{120}{\mathrm{x}}$

$\frac{120}{x}+\frac{80}{y}=100$.......(1)

Volume of ${\mathrm{O}}_2$ consumed $=\frac{40}{\mathrm{y}}(3\mathrm{y}+1)+\frac{60}{\mathrm{x}}\times \mathrm{x}$

Total initial volume of mixture $=100\mathrm{ml}+\mathrm{V}\mathrm{mL}$ of ${\mathrm{O}}_2$

Final volume of mixture $=80+60+{\mathrm{V}}^{\prime }\mathrm{mL}$ of ${\mathrm{O}}_2$

Volume of ${\mathrm{O}}_2$ consumed $=\mathrm{V}-{\mathrm{V}}^{\prime }$

$(100+\mathrm{V})-\left(80+60+{\mathrm{V}}^{\prime }\right)=160$

$\mathrm{V}-{\mathrm{V}}^{\prime }=200$

$\frac{40}{y}(3y+1)+\frac{60}{x}\times x=200$.........(2)

Solving (1) and (2); $\mathrm{x}=2$ and $\mathrm{y}=2$.

Volume of ${\mathrm{N}}_{\mathrm{x}}{\mathrm{H}}_{2\mathrm{x}}=60\mathrm{mL}$ and Volume of ${\mathrm{C}}_{\mathrm{y}}{\mathrm{H}}_{2\mathrm{y}+2}=40\mathrm{mL}$.