Calculate the amount of carbon dioxide that could be produced when
$(i)$ $1$ mole of carbon is burnt in air.
$(ii)$ $1$ mole of carbon is burnt in $16 \ g$ of dioxygen.
$(iii)$ $2$ moles of carbon are burnt in $16 \ g$ of dioxygen.
$(i)$ $1$ mole of carbon is burnt in air.
$(ii)$ $1$ mole of carbon is burnt in $16 \ g$ of dioxygen.
$(iii)$ $2$ moles of carbon are burnt in $16 \ g$ of dioxygen.
(A) The balanced chemical equation for the combustion of carbon is:
$C(s) + O_2(g) \rightarrow CO_2(g)$
$(i)$ From the balanced equation,$1$ mole of $C$ reacts with $1$ mole of $O_2$ to produce $1$ mole of $CO_2$ $(44 \ g)$. Since air is in excess,$1$ mole of $CO_2$ $(44 \ g)$ is produced.
$(ii)$ Molar mass of $O_2 = 32 \ g/mol$. Given $16 \ g$ of $O_2 = 0.5$ mole. Since $1$ mole of $C$ requires $1$ mole of $O_2$,$O_2$ is the limiting reactant. Thus,$0.5$ mole of $O_2$ produces $0.5$ mole of $CO_2$,which is $0.5 \times 44 = 22 \ g$.
$(iii)$ Here,$16 \ g$ ($0.5$ mole) of $O_2$ is again the limiting reactant. It will react with $0.5$ mole of $C$ to produce $0.5$ mole of $CO_2$,which is $22 \ g$.
$C(s) + O_2(g) \rightarrow CO_2(g)$
$(i)$ From the balanced equation,$1$ mole of $C$ reacts with $1$ mole of $O_2$ to produce $1$ mole of $CO_2$ $(44 \ g)$. Since air is in excess,$1$ mole of $CO_2$ $(44 \ g)$ is produced.
$(ii)$ Molar mass of $O_2 = 32 \ g/mol$. Given $16 \ g$ of $O_2 = 0.5$ mole. Since $1$ mole of $C$ requires $1$ mole of $O_2$,$O_2$ is the limiting reactant. Thus,$0.5$ mole of $O_2$ produces $0.5$ mole of $CO_2$,which is $0.5 \times 44 = 22 \ g$.
$(iii)$ Here,$16 \ g$ ($0.5$ mole) of $O_2$ is again the limiting reactant. It will react with $0.5$ mole of $C$ to produce $0.5$ mole of $CO_2$,which is $22 \ g$.
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