(i) $\underset{\text{1 mole (12g)}}{\mathrm{C}} + \underset{\text{1 mole (32g)}}{\mathrm{O}_2}$ → $\underset{\text{1 mole (44g)}}{\mathrm{CO}_2}$ Hence, 1 mole of C produces 44 g of $\mathrm{CO}_2$. (ii) $\underset{12g}{\mathrm{C}}+\underset{32g}{\mathrm{O}_2}$ → $\underset{44g}{\mathrm{CO}_2}$ Here, $\mathrm{O}_2$ is the limiting reagent. Since 32 g of $\mathrm{O}_2$ reacts with C to produce 44 g of $\mathrm{CO}_2$ Since 16 g of $\mathrm{O}_2$ reacts with C to produce $\frac{44}{32} \times 16$ = 22 g of $\mathrm{CO}_2$ (iii) $\underset{24g}{2\mathrm{C}}+\underset{64g}{2\mathrm{O}_2}$ → $\underset{88g}{2\mathrm{CO}_2}$ Here, $\mathrm{O}_2$ is the limiting reagent. Since 64 g of $\mathrm{O}_2$ reacts with C to produce 88 g of $\mathrm{CO}_2$ Since 16 g of $\mathrm{O}_2$ reacts with C to produce