A closed 10 , L vessel contains 1 , L water g | vedclass

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(B) At $STP$,$1 \, L$ of water gas contains $1:1$ ratio of $CO$ and $H_2$ gases. Thus,$V_{CO} = 0.5 \, L$ and $V_{H_2} = 0.5 \, L$. Volume of $O_2$ in

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$0.022$

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(B) At $STP$,$1 \, L$ of water gas contains $1:1$ ratio of $CO$ and $H_2$ gases. Thus,$V_{CO} = 0.5 \, L$ and $V_{H_2} = 0.5 \, L$. Volume of $O_2$ in $9 \, L$ of air $= 9 	imes 0.20 = 1.8 \, L$. Upon ignition,$CO$ reacts with $O_2$ according to the equation: $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)}$. Since $CO$ is the limiting reagent ($0.5 \, L$ of $CO$ requires $0.25 \, L$ of $O_2$),the volume of $CO_2$ produced is equal to the volume of $CO$ consumed,which is $0.5 \, L$. Number of moles of $CO_2 = \frac{	ext{Volume at } STP}{22.4 \, L/mol} = \frac{0.5}{22.4} \approx 0.022 \, mol$.

$A$ closed $10 \, L$ vessel contains $1 \, L$ water gas $(1:1 \, CO : H_2)$ and $9 \, L$ air ($20 \% \, O_2$ by volume) at $STP$. The contents of the vessel are ignited. The number of moles of $CO_2$ in the vessel is closest to $.... \, mol$.

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