Calculate the amount of carbon dioxide 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.
(N/A) The balanced chemical equation is: $C(s) + O_2(g) \rightarrow CO_2(g)$
Stoichiometry: $1 \ mol \ C$ reacts with $1 \ mol \ (32 \ g)$ of $O_2$ to produce $1 \ mol \ (44 \ g)$ of $CO_2$.
$(i)$ Since $1 \ mol$ of $C$ is burnt in excess air,it produces $44 \ g$ of $CO_2$.
$(ii)$ Here,$1 \ mol \ C$ requires $32 \ g$ of $O_2$. Since only $16 \ g$ of $O_2$ is provided,$O_2$ is the limiting reagent. Thus,$16 \ g$ of $O_2$ will produce $(44 \ g \ CO_2 / 32 \ g \ O_2) \times 16 \ g \ O_2 = 22 \ g$ of $CO_2$.
$(iii)$ Here,$2 \ moles \ C$ are provided with $16 \ g \ (0.5 \ mol)$ of $O_2$. $O_2$ is again the limiting reagent. Thus,$16 \ g$ of $O_2$ will produce $22 \ g$ of $CO_2$.
Stoichiometry: $1 \ mol \ C$ reacts with $1 \ mol \ (32 \ g)$ of $O_2$ to produce $1 \ mol \ (44 \ g)$ of $CO_2$.
$(i)$ Since $1 \ mol$ of $C$ is burnt in excess air,it produces $44 \ g$ of $CO_2$.
$(ii)$ Here,$1 \ mol \ C$ requires $32 \ g$ of $O_2$. Since only $16 \ g$ of $O_2$ is provided,$O_2$ is the limiting reagent. Thus,$16 \ g$ of $O_2$ will produce $(44 \ g \ CO_2 / 32 \ g \ O_2) \times 16 \ g \ O_2 = 22 \ g$ of $CO_2$.
$(iii)$ Here,$2 \ moles \ C$ are provided with $16 \ g \ (0.5 \ mol)$ of $O_2$. $O_2$ is again the limiting reagent. Thus,$16 \ g$ of $O_2$ will produce $22 \ g$ of $CO_2$.
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