Chemistry - Ideal Gas Question with Solution | TestHub

${\mathrm{CO}}_2+\mathrm{C}⟶2\mathrm{CO}$ Stoichoimetry ratio is $1:2$ AT STP, $P=1\mathrm{atm},T=273\mathrm{K},R=0.0821$ Initial moles of ${\mathrm{CO}}_2;n({\mathrm{CO}}_2$ initial $)=\frac{PV}{RT}$ $=\frac{1\times 0.5}{0.0821\times 273}=0.022\mathrm{mole}$ In final mixture no. of moles; $n({\mathrm{CO}}_2/\mathrm{CO}$ mixture) $$=\frac{1\times 0.7}{0.0821\times 273}=0.031$$ Increase in volume is by $=0.031-0.022$ $=0.009$ mole of gas Final no. of moles of $\mathrm{CO}$ i.e. $n_{\text{(CO final) }}$ $$\begin{array}{rl}& n_{\text{(CO final) }}=2n_{\left({\mathrm{CO}}_2\text{ initial }\right)}-n_{({\mathrm{CO}}_2\text{ final) }}\\ & =2(0.022-{\mathrm{n}}_{\left({\mathrm{CO}}_2\text{ final }\right)}\\ & n_{\text{(CO final) }}=0.044-2n_{\text{(CO }}\text{ final })\\ & \therefore \text{ Now, }{n}_{\text{(CO final) }}+n_{\text{(CO final) }}=0.031\\ & n_{({\mathrm{CO}}_2\text{ final) }}=0.031-n_{\text{(CO final) })}\dots \text{ (iii) }\end{array}$$ Substituting (ii) in eq. (i) $$\begin{array}{rl}& n_{\text{(CO final) }}=0.044-2[0.031-n_{\text{(CO final) }}\\ & n_{\text{(CO final) }}=0.044-0.062+2n_{\text{(CO final })}\end{array}$$ $$n_{(\mathrm{CO}\text{ final) }}=0.018\mathrm{mol}.$$ Volume of $\mathrm{CO}=V=\frac{nRT}{P}=\frac{0.018\times 0.0821\times 273}{1}$ $=0.40$ Litre and volume of ${\mathrm{CO}}_2=0.7$ litre $-0.4$ litre $=0.3$ litre $$\therefore {\mathrm{CO}}_2=300\mathrm{mL},\mathrm{CO}=400\mathrm{mL}$$