When CO_ 2(~g) is passed over red hot coke it partially gets...

\mathrm{CO}_2+\mathrm{C} \longrightarrow 2 \mathrm{CO}$Stoichoimetry ratio is$1: 2$AT STP,$P=1 \mathrm{ atm}, T=273 \mathrm{ K}, R=0.0821$Initial moles of$\mathrm{CO}_2 ; n\left(\mathrm{CO}_2\right.$initial$)=\frac{P V}{R T}=\frac{1 \times 0.5}{0.0821 \times 273}=0.022 \mathrm{ mole}$In final mixture no. of moles;$n\left(\mathrm{CO}_2 / \mathrm{CO}\right.$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{aligned}& n_{\text{(\text{ CO }\text{ final }) }}=2 n_{\left(\mathrm{CO}_{2} \text{ initial }\right)}-n_{\left(\mathrm{CO}}_{2}\right.\text{ final) }} \\ & \quad=2\left(0.022-\mathrm{n}_{\left(\mathrm{CO}_2 \text{ final }\right)}\right.\\ & n_{\text{(CO final) }}=0.044-2 n_{\text{(CO }} \text{ final }\right)\\ & \therefore \text{ Now, } n_{\text{(CO final) }}+n_{\text{(CO final) }}=0.031 \\ & n_{\left(\mathrm{CO}_{2}\right. \text{ final) }}=0.031-n_{\text{(CO final) })} \ldots \text{ (iii) } \end{aligned}Substituting (ii) in eq. (i) $$\begin{aligned} & n_{\text{(CO final) }}=0.044-2\left[0.031-n_{\text{(CO final) }}\right. \\ & n_{\text{(CO final) }}=0.044-0.062+2 n_{\text{(CO final })} \end{aligned}$$ n_{(\mathrm{CO} \text{ final) }}=0.018 \mathrm{ mol} .Volume of \mmathrm{CO}=V=\frac{n R T}{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}