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Question
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.
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Solution
(i) 1 mole of carbon is burnt in air.
C+O2→CO2
1 mole of C produces 1 mole of CO2 which corresponds to 44 g of CO2.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
16 g of dioxygen corresponds to 1632=0.5 moles.
Here, dioxygen is the limiting reagent.
It will produce 0.5 moles of CO2 which corresponds to 0.5×44=22 g of CO2.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
16 g of dioxygen corresponds to 1632=0.5 moles.
Here, dioxygen is the limiting reagent.
It will produce 0.5 moles of CO2 which corresponds to 0.5×44=22 g of CO2.
C+O2→CO2
1 mole of C produces 1 mole of CO2 which corresponds to 44 g of CO2.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
16 g of dioxygen corresponds to 1632=0.5 moles.
Here, dioxygen is the limiting reagent.
It will produce 0.5 moles of CO2 which corresponds to 0.5×44=22 g of CO2.
16 g of dioxygen corresponds to 1632=0.5 moles.
Here, dioxygen is the limiting reagent.
It will produce 0.5 moles of CO2 which corresponds to 0.5×44=22 g of CO2.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
16 g of dioxygen corresponds to 1632=0.5 moles.
Here, dioxygen is the limiting reagent.
It will produce 0.5 moles of CO2 which corresponds to 0.5×44=22 g of CO2.
16 g of dioxygen corresponds to 1632=0.5 moles.
Here, dioxygen is the limiting reagent.
It will produce 0.5 moles of CO2 which corresponds to 0.5×44=22 g of CO2.
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