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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 \(16 \mathrm{~g}\) of dioxygen.
(iii) 2 moles of carbon are burnt in \(16 \mathrm{~g}\) of dioxygen.
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in \(16 \mathrm{~g}\) of dioxygen.
(iii) 2 moles of carbon are burnt in \(16 \mathrm{~g}\) of dioxygen.
Solution:
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Verified Answer
The balanced equation for the combustion of carbon in dioxygen/air is
(i) In air, combustion is complete. Therefore, \(\mathrm{CO}_2\) produced from the combustion of 1 mole of carbon \(=44 \mathrm{~g}\).
(ii) Since only \(16 \mathrm{~g}\) of dioxygen is available, hence it can combine only with \(0.5\) mole of carbon, i.e., dioxygen is the limiting reactant. Hence, \(\mathrm{CO}_2\) produced \(=22 \mathrm{~g}\).
(iii) In this case dioxygen again is the limiting reactant. \(16 \mathrm{~g}\) of dioxygen can combine only with \(0.5\) mole of carbon. \(\mathrm{CO}_2\) produced again is equal to \(22 \mathrm{~g}\).
(i) In air, combustion is complete. Therefore, \(\mathrm{CO}_2\) produced from the combustion of 1 mole of carbon \(=44 \mathrm{~g}\).
(ii) Since only \(16 \mathrm{~g}\) of dioxygen is available, hence it can combine only with \(0.5\) mole of carbon, i.e., dioxygen is the limiting reactant. Hence, \(\mathrm{CO}_2\) produced \(=22 \mathrm{~g}\).
(iii) In this case dioxygen again is the limiting reactant. \(16 \mathrm{~g}\) of dioxygen can combine only with \(0.5\) mole of carbon. \(\mathrm{CO}_2\) produced again is equal to \(22 \mathrm{~g}\).
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