Question

In the reaction:
$4{\mathbf{A}} + 2{\mathbf{B}} + 3{\mathbf{C}} \to {{\mathbf{A}}_4}{{\mathbf{B}}_2}{{\mathbf{C}}_3}$
What will be the number of moles of product formed, starting from one mole of A, 0.6 mole of B and 0.72 mole of C?

Answer 1

In the reactions where more than one reactant is involved, the amount of product is formed in proportion to the amount of the reactant which is entirely consumed. Thus, we can define the limiting reagent as the reactant that is entirely consumed during the course of a chemical reaction.

Complete Step by step answer:
Here the question is direct and simple. We just have to find the limiting reagent of this reaction and it will lead into our solution. So, in order to find the limiting reagent let us apply some stoichiometry.
Here, it is given that 4 moles of A, 2 moles B and 3 moles of C combine to give 1 mole of ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$. We can cut short this into and say,
4 mole of A requires 2 moles of B in order to produce ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$.
1 mole of A requires $\dfrac{2}{4}$ moles of B
1 mole of A requires 0.5 moles of B
In question it is given that we have 0.6 moles of B. Here, we only require 0.5 moles to produce ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$. Therefore we can say it is not the limiting reagent.
Now, let’s look into C.
4 moles of A requires 3 moles C in order to produce ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$.
1 mole of A requires $\dfrac{3}{4}$ moles of B
1 mole of A requires 0.75 moles of B
But in question it is given that we have only 0.72 moles of C. Therefore the number of moles required to produce ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$ by C is less. Therefore we take C as the limiting reagent.
So, moles of product now depends on the moles of C used.
Therefore, 3 moles of C produces 1 moles of ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$
1 mole of C produces $\dfrac{1}{3}$ moles of ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$
0.72 moles of C produces $\dfrac{{0.72}}{3}$ moles of ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$
i.e., 0.14 moles of ${{\text{A}}_4}{{\text{B}}_2}{{\text{C}}_3}$ is formed.

Hence we get a 0.14 moles of product.

Note: The most important thing to notice here is that balancing the reaction. Here the given reaction is already balanced hence, we can apply the stoichiometry directly. But when the reaction is given as a word we should not forget to balance the reaction. Otherwise the entire problem will be wrong.