Answer:
(i)
\[\underset{1\,\text{mol}}{\mathop{C(s)}}\,+{{O}_{2}}(g)\to
\underset{1\,\text{mol}}{\mathop{C{{O}_{2}}(g)}}\,\]
Above reaction shows that 1 \[mol\] carbon, when burnt in excess
air will give 1 \[mol\]\[C{{O}_{2}}.\]
(ii) (1 \[mol\]
carbon = 12 g carbon) and 16 g of dioxygen are allowed to react, according to
the following reaction:
\[\underset{\underset{1\times
12g}{\mathop{1\,\text{mol}}}\,}{\mathop{C(s)}}\,+\underset{\underset{1\times
\,32g}{\mathop{1\,\text{mol}}}\,}{\mathop{{{O}_{2}}(g)}}\,\to
\underset{\underset{1\times
\,44g}{\mathop{1\text{mol}}}\,}{\mathop{C{{O}_{2}}(g)}}\,\]
Case I: Let carbon is completely consumed in the
reaction:
12 g carbon on
complete consumption will give 44 g \[C{{O}_{2}}\]
Case II: Let oxygen is completely consumed in the
reaction:
\[\because \] \[32g\]oxygen
gives 44 g \[C{{O}_{2}}\]
\[\therefore
\] 16 g oxygen will give 22 g \[C{{O}_{2}}\]
Since,
oxygen on complete consumption gives least amount of product (\[C{{O}_{2}}\])
hence oxygen is limiting reactant and hence 22 g will be actually formed in the
reaction.
(iii) 2 \[mol\]
carbon i.e., 24 g carbon is allowed to react with 16 g dioxygen.
\[\underset{\underset{1\times
\,12g}{\mathop{1\,\text{mol}}}\,}{\mathop{C(s)}}\,+\underset{\underset{1\,\times
\,32g}{\mathop{1\,\text{mol}}}\,}{\mathop{{{O}_{2}}(g)}}\,\to
\underset{\underset{1\times \,44g}{\mathop{1\,\text{mol}}}\,}{\mathop{C{{O}_{2}}(g)}}\,\]
Case I: Let carbon is completely consumed in the
reaction:
\[\because \]12 g
carbon gives 44 g \[C{{O}_{2}}\]
\[\therefore \]24 g
carbon will give 88 g \[C{{O}_{2}}\]
Case II: Let oxygen is completely consumed in the
reaction:
\[\because \]32 g
oxygen gives 44 g \[C{{O}_{2}}\]
\[\therefore \]16 g
oxygen will give 22 g \[C{{O}_{2}}\]
Since, oxygen on complete consumption gives least amount of
product (\[C{{O}_{2}}\]) thus, oxygen is limiting reagent and hence 22 g \[C{{O}_{2}}\]
will be actually formed in the reaction.
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