Amount of CH₃CI that can be prepared from 20 g of CH₄ and 10 g of Cl₂ by the following reaction, is :

${\mathrm{CH}}_4+{\mathrm{Cl}}_2⟶{\mathrm{CH}}_3\mathrm{Cl}+\mathrm{HCl}$, (assume that no other reaction is taking place)

${\mathrm{CH}}_4+\mathrm{C}{\mathrm{l}}_2\to {\mathrm{CH}}_3\mathrm{Cl}+\mathrm{HCl}$

One mole of methane reacts with one mole of chlorine gas. 20 g of methane has 1.25 moles of methane and 10 g of Cl₂ has 0.3 moles of Cl₂. This shows that Cl₂ is the limiting reagent and decides the quantity of the product forms in the reaction.

One mole of Cl₂ forms one mole of CH₃Cl, therefore 0.3 moles of Cl₂ forms 0.3 moles of CH₃Cl. Amount of CH₃Cl is:

$0.3 \mathrm{mol}=\frac{\mathrm{Mass} \mathrm{of} {\mathrm{CH}}_3\mathrm{Cl}}{50.5 \mathrm{g}/\mathrm{mol}}=15.15 \mathrm{g}$