When 1 L of CO_2 gas is heated with graphite | vedclass

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(C) The chemical reaction is: $CO_2(g) + C(s) \rightarrow 2CO(g)$.Let the initial volume of $CO_2$ be $1 \ L$.Let $x \ L$ be the volume of $CO_2$ that

Vedclass · Accepted Answer

$1/22.4$

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(C) The chemical reaction is: $CO_2(g) + C(s) ightarrow 2CO(g)$.Let the initial volume of $CO_2$ be $1 \ L$.Let $x \ L$ be the volume of $CO_2$ that reacts.According to the stoichiometry,$x \ L$ of $CO_2$ produces $2x \ L$ of $CO$.The remaining volume of $CO_2$ is $(1 - x) \ L$.The total volume of the gas mixture at $STP$ is $(1 - x) + 2x = 1 + x$.Given that the total volume is $1.5 \ L$,we have $1 + x = 1.5$,so $x = 0.5 \ L$.The volume of $CO$ produced is $2x = 2 	imes 0.5 = 1 \ L$.The number of moles of $CO$ at $STP$ is $\frac{	ext{Volume}}{22.4} = \frac{1}{22.4}$ moles.

When $1 \ L$ of $CO_2$ gas is heated with graphite,the volume of the resulting gas mixture at $STP$ is $1.5 \ L$. Calculate the number of moles of $CO$ gas produced at $STP$.

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