Question

In the reaction $4A + 2B + 3C \to {A_4}{B_2}{C_3}$ , the numbers of moles of products formed will be_____, if starting material from $2$ moles of $A$ , $1.2$ moles of $B$ and $1.44$ moles of $C.$
$\left( A \right)$ $0.48$
$\left( B \right)$ $0.3$
$\left( C \right)$ $0.6$
$\left( D \right)$ $41$

Answer 1

For any balanced chemical reaction, whole number (coefficients) are used to show the quantities (generally in moles) of both the reactant and products, for example in the given reaction $4$ moles of A, $2$ moles of and $3$ moles of $C$ reacts to produce one mole of product.

Complete step by step answer:
We have the given equation, $4A + 2B + 3C \to {A_4}{B_2}{C_3}$
From the reaction, we may conclude that Number of moles of element $B$ required to react with $4$ moles of element $A = 2$
Or, Number of moles of element $B$ required to react with $2$ moles of element $A = \dfrac{2}{4} \times 2$
$\Rightarrow$ $= 1$
Thus, $1$ mole of element $B$ required to react with $2$ moles of element $A$
In the question given amount of element $B$ $= 1.2$ moles
And, Number of moles of element $C$ required to react with $4$ moles of element $A$ $= 3$
Or, Number of moles of element $C$ required to react with $2$ moles of element $A$ $= \dfrac{3}{4} \times 2$
$\Rightarrow$ $= 1.5$
Thus, $1.5$ moles of element $C$ required to react with $2$ moles of element $A$
In the question given the amount of element $C$ $= 1.44$
Therefore, the amount of element $C$ is less than the amount required to react with element $A$, $C$ is taken as a limiting reagent.

Now, Number of moles of formed when $3$ moles of element $C$ reacts $= 1$ mole
Or, Number of moles product formed when $1.44$ moles of element $C$ reacts $= \dfrac{1}{3} \times 1.44$
$\Rightarrow$ $= 0.48$
Hence, the product formed is $0.48$ moles
So, the correct option is $A.$

Note:
A limiting reagent is defined as the in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.