Chemistry - Stoichiometry Question with Solution | TestHub

${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right){\mathrm{O}}_2\to {\mathrm{xCO}}_2+\frac{\mathrm{y}}{2}{\mathrm{H}}_2\mathrm{O}$

$\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=0.05$ moles

${\mathrm{n}}_{{\mathrm{CO}}_2}=0.05\mathrm{x}\Rightarrow 0.05\mathrm{x}\times 44=2.2$

$\Rightarrow \mathrm{x}=1$

&

${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=0.05\frac{\mathrm{y}}{2}\Rightarrow \frac{0.05\mathrm{y}}{2}\times 18=1.8$

$\Rightarrow \mathrm{y}=4$

So the reaction

${\mathrm{CH}}_4+2{\mathrm{O}}_2⟶{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$

for 0.05 mole ${\mathrm{CH}}_4,{\mathrm{n}}_{{\mathrm{O}}_2}$ reqd. $=2\times 0.05=0.1$ mole

$=2.24\mathrm{L}$ at 1 atm and 273 K

${\mathrm{w}}_{{\mathrm{CH}}_4}=0.05\times 16=0.8\mathrm{g}$