Stoichiometry
2.0 The Limiting Reagent
2.0 The Limiting Reagent
The reactions in which more than one reactants are reacting and if they are not present in the same molar ratio as the balanced equation requires, then one has to find out the limiting reagent i.e., the reagent which is completely consumed in the reaction.
All quantitative calculations are carried out with the help of limiting reagent only. Now, how to decide the limiting reagent of a reaction? Consider the following example.
Suppose you are given $5$ moles of $PbS$ and $18$ moles of ${H_2}{O_2}$ and you have to calculate the maximum amount of $PbS{O_4}$ being produced.
The balanced equation is:
$$\begin{equation} \begin{aligned} \quad \quad \quad \quad \quad \quad PbS + {\text{ }}4{H_2}{O_2} \to PbS{O_4} + {\text{ }}4{H_2}O \\ {\text{Initial moles 5 18}} \\\end{aligned} \end{equation} $$
Now since $1$ mole of $PbS$ reacts completely with $4$ moles of ${H_2}{O_2}$ to produce $1$ mole of $PbS{O_4}$. Therefore, $5$ moles of $PbS$ will react with $20$ moles of ${H_2}{O_2}$.
But since moles of ${H_2}{O_2}$ is only $18$.
So, ${H_2}{O_2}$ will be the reagent which will be consumed first, and hence ${H_2}{O_2}$ is the limiting reagent.
If $PbS$ have been the limiting reagent $5$ mole of it would have given $5$ moles of $PbS{O_4}$ but it is not so.
As one can see from the balanced chemical equation that $1$ mole of $PbS{O_4}$ is produced from $4$ moles of ${H_2}{O_2}$ So $18$ moles of ${H_2}{O_2}$ will produce $4.5$ moles of $PbS{O_4}$.
So, one can define limiting reagent in another way,
“The reagent producing the least number of moles of products is the limiting reagent”.
