Question

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 the gases increased to $700\, mL$. The composition of the gaseous mixture at $STP$ is

• A. $CO_2 = 300\, mL; CO=400 \,mL$
• B. $CO_2 = 0.0\, mL; CO=700 \,mL$
• C. $CO_2 = 200\, mL; CO=500 \,mL$
• D. $CO_2 = 350\, mL; CO=350 \,mL$

Answer 1

The Correct Option is A

$ {CO2+C->2CO}$ Stoichoimetry ratio is $1 : 2$ AT $STP, P= 1 $ atm, $r= 273 K, R = 0.0821$ Initial moles of $CO_2. n$($CO_2$ initial) $=\frac{PV}{RT}$ $=\frac{1\times0.5}{0.0821\times273}=0.022$ mole In final mixture no. of moles; $n$($CO_2/CO$ mixture) $=\frac{1\times0.5}{0.0821\times273}=0.031$ Increase in volume is by $= 0.031 ? 0.022$ $=0.009$ mole of gas Final no. of moles of $CO$ i.e. $n_{\left(CO\,final\right)}$ $n_{\left(CO\,final\right)}=2n_{\left(CO_2\,initial\right)}-n_{\left(CO_2\,final\right)}$ $=2\left(0.002-n_{\left(CO_2\,final\right)}\right)\,...\left(i\right)$ $n_{\left(CO\,final\right)}=0.044-2n_{\left(CO_2\,final\right)}\,...\left(ii\right)$ $\therefore$ Now, $n_{\left(CO\,final\right)}+n_{\left(CO\,final\right)}=0.031$ $n_{\left(CO_2\,final\right)}=0.031-n_{\left(CO\,final\right)}\,...\left(iii\right)$ Substituting $\left(ii\right)$ in e $\left(i\right)$ $n_{\left(CO\,final\right)}=0.004-2\left[0.031-n_{\left(CO\,final\right)}\right]$ $n_{\left(CO\,final\right)}=0.044-0.062+2n_{\left(CO\,final\right)}$ $n_{\left(CO\,final\right)}=0.018\,mol.$ Volume of $CO=V=\frac{nRT}{P}=\frac{0.018\times0.0821\times273}{1}$ $=0.40\,Litre$ and volume of $CO_{2} = 0.7 litre - 0.4 \,litre$ $=0.3\,litre$ $\therefore CO_{2}=300\,mL, CO=400\,mL$