Assume carbon burns according to the followin | vedclass

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(C) The balanced chemical equation is: $2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}$Moles of $C = \frac{12 \ g}{12 \ g \ mol^{-1}} = 1 \ mol$.Moles of $

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$227$

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(C) The balanced chemical equation is: $2C_{(s)} + O_{2(g)} ightarrow 2CO_{(g)}$Moles of $C = \frac{12 \ g}{12 \ g \ mol^{-1}} = 1 \ mol$.Moles of $O_2 = \frac{48 \ g}{32 \ g \ mol^{-1}} = 1.5 \ mol$.According to the stoichiometry,$2 \ mol$ of $C$ requires $1 \ mol$ of $O_2$. Therefore,$1 \ mol$ of $C$ requires $0.5 \ mol$ of $O_2$.Since we have $1.5 \ mol$ of $O_2$,$C$ is the limiting reagent.From the equation,$2 \ mol$ of $C$ produces $2 \ mol$ of $CO$,so $1 \ mol$ of $C$ produces $1 \ mol$ of $CO$.Volume of $1 \ mol$ of $CO$ at $STP$ = $22.7 \ L = 227 	imes 10^{-1} \ L$.

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