When CO_2( ~g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 ~L of CO_2(g) over red hot coke, the total volume of…

$\mathrm{CO}_2+\mathrm{C} \longrightarrow 2 \mathrm{CO}$CO2​+C⟶2CO Stoichoimetry ratio is $1: 2$1:2 AT STP, $P=1 \mathrm{~atm}, T=273 \mathrm{~K}, R=0.0821$P=1 atm,T=273 K,R=0.0821 Initial moles of $\mathrm{CO}_2 ; n\left(\mathrm{CO}_2\right.$CO2​;n(CO2​ initial $)=\frac{P V}{R T}$)=RTPV​ $=\frac{1 \times 0.5}{0.0821 \times 273}=0.022 \mathrm{~mole}$=0.0821×2731×0.5​=0.022 mole In final mixture no. of moles; $n\left(\mathrm{CO}_2 / \mathrm{CO}\right.$n(CO2​/CO mixture) $$=\frac{1 \times 0.7}{0.0821 \times 273}=0.031$$=0.0821×2731×0.7​=0.031 Increase in volume is by $=0.031-0.022$=0.031−0.022 $=0.009$=0.009 mole of gas Final no. of moles of $\mathrm{CO}$CO i.e. $n_{\text {(CO final) }}$n(CO final) ​ $$\begin{aligned} & n_{\text {(CO final) }}=2 n_{\left(\mathrm{CO}_2 \text { initial }\right)}-n_{\left(\mathrm{CO}_2\right. \text { final) }} \\ & \quad=2\left(0.022-\mathrm{n}_{\left(\mathrm{CO}_2 \text { final }\right)}\right. \\ & \left.n_{\text {(CO final) }}=0.044-2 n_{\text {(CO }} \text { final }\right) \\ & \therefore \text { Now, } n_{\text {(CO final) }}+n_{\text {(CO final) }}=0.031 \\ & n_{\left(\mathrm{CO}_2\right. \text { final) }}=0.031-n_{\text {(CO final) })} \ldots \text { (iii) } \end{aligned}$$​n(CO final) ​=2n(CO2​ initial )​−n(CO2​ final) ​=2(0.022−n(CO2​ final )​n(CO final) ​=0.044−2n(CO ​ final )∴ Now, n(CO final) ​+n(CO final) ​=0.031n(CO2​ final) ​=0.031−n(CO final) )​… (iii) ​ 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(CO final) ​=0.044−2[0.031−n(CO final) ​n(CO final) ​=0.044−0.062+2n(CO final )​​ $$n_{(\mathrm{CO} \text { final) }}=0.018 \mathrm{~mol} .$$n(CO final) ​=0.018 mol. Volume of $\mathrm{CO}=V=\frac{n R T}{P}=\frac{0.018 \times 0.0821 \times 273}{1}$CO=V=PnRT​=10.018×0.0821×273​ $=0.40$=0.40 Litre and volume of $\mathrm{CO}_2=0.7$CO2​=0.7 litre $-0.4$−0.4 litre $=0.3$=0.3 litre $$\therefore \mathrm{CO}_2=300 \mathrm{~mL}, \mathrm{CO}=400 \mathrm{~mL}$$∴CO2​=300 mL,CO=400 mL