• # question_answer 12) 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) $\underset{1\,\text{mol}}{\mathop{C(s)}}\,+{{O}_{2}}(g)\to \underset{1\,\text{mol}}{\mathop{C{{O}_{2}}(g)}}\,$ 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.