The balanced reaction of combustion of carbon can be written as:
\( \underset{(1 mole)}{C_{(s)}} + \underset{\underset{(32 g)}{(1 mole)}}{O_{2_{(g)}}} → \underset{\underset{(44 g)}{(1 mole)}}{CO_{2_{(g)}}}\)
(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide.
(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.
(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.
LIST I | LIST II |
---|---|
A. 88 g of CO2 | I. 2 mol |
B. 6.023 × 1023 molecules of H2O | II. 1 mol |
C. 96 g of O2 | III. 6.023 × 1023 molecules |
D. 1 mol of any gas | IV. 3 mol |