When {CO_2}_{(g)} is passed over hot coke,it | vedclass

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(D) The chemical reaction is: ${CO_2}_{(g)} + C_{(s)} \rightarrow 2CO_{(g)}$.Let the volume of ${CO_2}$ reacted be $x \, mL$.According to the stoichio

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The chemical reaction is: ${CO_2}_{(g)} + C_{(s)} ightarrow 2CO_{(g)}$.Let the volume of ${CO_2}$ reacted be $x \, mL$.According to the stoichiometry,$x \, mL$ of ${CO_2}$ produces $2x \, mL$ of ${CO}$.The initial volume of ${CO_2}$ is $500 \, mL$.The volume of ${CO_2}$ remaining is $(500 - x) \, mL$.The volume of ${CO}$ produced is $2x \, mL$.The total volume of the final mixture is $(500 - x) + 2x = 500 + x \, mL$.Given that the total volume increases by $700 \, mL$ from the initial $500 \, mL$,the final volume is $500 + 700 = 1200 \, mL$.Equating the two: $500 + x = 1200$,so $x = 700 \, mL$.However,we only had $500 \, mL$ of ${CO_2}$ initially.If all $500 \, mL$ of ${CO_2}$ reacts $(x = 500)$,the final volume is $500 + 500 = 1000 \, mL$.Re-evaluating the problem statement: If the volume increases by $200 \, mL$ (total $700 \, mL$),then $x = 200 \, mL$.Remaining ${CO_2} = 500 - 200 = 300 \, mL$ and $CO = 2 	imes 200 = 400 \, mL$.

When ${CO_2}_{(g)}$ is passed over hot coke,it is reduced to ${CO}_{(g)}$. When $0.5 \, L$ of ${CO_2}_{(g)}$ is passed over hot coke,the total volume of the gas increases by $700 \, mL$ at $STP$. The composition of the resulting gaseous mixture is:

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