When CO2(g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 L of CO2(g) over red hot coke, the total volume of the gases increased to 700 mL. The composition of the gaseous mixture at STP is

Question

When CO2(g) is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5 L of CO2(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) CO2 = 300 mL; CO=400 mL (B) CO2 = 0.0 mL; CO=700 mL (C) CO2 = 200 mL; CO=500 mL (D) CO2 = 350 mL; CO=350 mL

Tardigrade Admin · Accepted Answer

CO2+C arrow 2CO Stoichoimetry ratio is 1: 2 AT STP, P= 1 atm, r= 273 K, R = 0.0821 Initial moles of CO2. n(CO2 initial) =(PV/RT) =(1×0.5/0.0821×273)=0.022 mole In final mixture no. of moles; n(CO2/CO mixture) =(1×0.5/0.0821×273)=0.031 Increase in volume is by = 0.031 â 0.022 =0.009 mole of gas Final no. of moles of CO i.e. n(CO final) n(CO final)=2n(CO2 initial)-n(CO2 final) =2(0.002-n(CO2 final)) ...(i) n(CO final)=0.044-2n(CO2 final) ...(ii) ∴ Now, n(CO final)+n(CO final)=0.031 n(CO2 final)=0.031-n(CO final) ...(iii) Substituting (ii) in eq. (i) n(CO final)=0.004-2[0.031-n(CO final)] n(CO final)=0.044-0.062+2n(CO final) n(CO final)=0.018 mol. Volume of CO=V=(nRT/P)=(0.018×0.0821×273/1) =0.40 Litre and volume of CO2 = 0.7 litre - 0.4 litre =0.3 litre ∴ CO2=300 mL, CO=400 mL

Q. When $C O_{2 (g)}$ is passed over red hot coke it partially gets reduced to $CO (g)$ . Upon passing $0.5 L$ of $C O_{2 (g)}$ over red hot coke, the total volume of the gases increased to $700 m L$ . The composition of the gaseous mixture at $STP$ is

$C O_{2} = 300 m L; CO = 400 m L$

$C O_{2} = 0.0 m L; CO = 700 m L$

$C O_{2} = 200 m L; CO = 500 m L$

$C O_{2} = 350 m L; CO = 350 m L$

Q. When CO2(g)​ is passed over red hot coke it partially gets reduced to CO(g). Upon passing 0.5L of CO2(g)​ over red hot coke, the total volume of the gases increased to 700mL. The composition of the gaseous mixture at STP is

CO2​=300mL;CO=400mL

CO2​=0.0mL;CO=700mL

CO2​=200mL;CO=500mL

CO2​=350mL;CO=350mL

Q. When $C O_{2 (g)}$ is passed over red hot coke it partially gets reduced to $CO (g)$ . Upon passing $0.5 L$ of $C O_{2 (g)}$ over red hot coke, the total volume of the gases increased to $700 m L$ . The composition of the gaseous mixture at $STP$ is

$C O_{2} = 300 m L; CO = 400 m L$

$C O_{2} = 0.0 m L; CO = 700 m L$

$C O_{2} = 200 m L; CO = 500 m L$

$C O_{2} = 350 m L; CO = 350 m L$