When 1 dm^{3} of CO_{2} gas is passed over h | vedclass

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Question

(D) The chemical reaction is: $CO_{2}(g) + C(s) \rightarrow 2CO(g)$.Let the initial volume of $CO_{2}$ be $1 \ dm^{3}$.Let $x$ be the volume of $CO_{2

Vedclass · Accepted Answer

$0.8 \ dm^{3}$ of $CO$, $0.6 \ dm^{3}$ of $CO_{2}$

Vedclass · Answer

(D) The chemical reaction is: $CO_{2}(g) + C(s) \rightarrow 2CO(g)$.Let the initial volume of $CO_{2}$ be $1 \ dm^{3}$.Let $x$ be the volume of $CO_{2}$ that reacts with carbon.According to the stoichiometry of the reaction, $1 \ mole$ of $CO_{2}$ produces $2 \ moles$ of $CO$. Thus, $x \ dm^{3}$ of $CO_{2}$ will produce $2x \ dm^{3}$ of $CO$.The remaining volume of $CO_{2}$ is $(1 - x) \ dm^{3}$.The total volume of the gaseous mixture is the sum of the remaining $CO_{2}$ and the produced $CO$: $(1 - x) + 2x = 1 + x$.Given that the final volume is $1.4 \ dm^{3}$, we have $1 + x = 1.4$, which gives $x = 0.4 \ dm^{3}$.Therefore, the volume of $CO$ produced is $2x = 2(0.4) = 0.8 \ dm^{3}$.The volume of remaining $CO_{2}$ is $1 - 0.4 = 0.6 \ dm^{3}$.

When $1 \ dm^{3}$ of $CO_{2}$ gas is passed over hot coke, the volume of the gaseous mixture after the complete reaction at $STP$ becomes $1.4 \ dm^{3}$. The composition of the gaseous mixture at $STP$ is:

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