When CO_2,(g) is passed over red hot coke,it | vedclass

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(A) The chemical reaction is: $CO_2(g) + C(s) 	o 2CO(g)$Let the volume of $CO_2$ reacted be $x\,mL$.According to the stoichiometry,$x\,mL$ of $CO_2$

Vedclass · Accepted Answer

$CO_2 = 300\,mL;\,CO = 400\,mL$

Vedclass · Answer

(A) The chemical reaction is: $CO_2(g) + C(s) 	o 2CO(g)$Let the volume of $CO_2$ reacted be $x\,mL$.According to the stoichiometry,$x\,mL$ of $CO_2$ produces $2x\,mL$ of $CO$.Initial volume of $CO_2 = 500\,mL$.Remaining volume of $CO_2 = (500 - x)\,mL$.Volume of $CO$ produced = $2x\,mL$.Total final volume = $(500 - x) + 2x = 500 + x$.Given that the total final volume is $700\,mL$,so $500 + x = 700$,which gives $x = 200\,mL$.Volume of $CO_2$ remaining = $500 - 200 = 300\,mL$.Volume of $CO$ produced = $2 	imes 200 = 400\,mL$.Thus,the composition is $CO_2 = 300\,mL$ and $CO = 400\,mL$.

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

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