Question

In the reaction $4A + 2B + 3C \to {A_4}{B_2}{C_3}$ ​, the number of moles of products formed will be ______, if starting from $2$ moles of A, $1.2$ moles of B and $1.44\;$moles of C.
(A) $0.48$
(B) $0.3$
(C) $0.6$
(D) $41$

Answer 1

The number of moles of product that will be formed from the given reactants in the reaction depends on the reactant that will act as the limiting reagent. So, we will first find the limiting reagent then we will calculate the total number of moles of product formed.

Complete step by step answer:
The given reaction is:
$4A + 2B + 3C \to {A_4}{B_2}{C_3}$
From the reaction, we can deduce that,
Number of moles of A required to form $1$ mole of ${A_4}{B_2}{C_3} = 4$
Number of moles of B required to form $1$ mole of ${A_4}{B_2}{C_3} = 2$
Number of moles of C required to form $1$ mole of ${A_4}{B_2}{C_3} = 3$
Now, we are given, Total number of moles of A $= 2$
Total number of moles of B $= 1.2$
And, Total number of moles of C $= 1.44$
Now we will analyse this data to find the limiting reagent in the reaction.

| Moles of A| Moles of B| Moles of C| Moles of ${A_4}{B_2}{C_3}$
---|---|---|---|---
Initial| $4$| $2$| $3$| $1$
Final| $2$| $1$| $\dfrac{3}{2} = 1.5$| $\dfrac{1}{2} = 0.5$

As you can see from the table, the moles of C required is less than the given amount, Hence, it will be the limiting reagent. Now we will calculate the number of moles of product formed from $1.44$ moles of reactant C.
Number of moles of product formed when $3$ mole of C reacts $= 1$
Number of moles of product formed when $1$ mole of C reacts $= \dfrac{1}{3}$
Number of moles of product formed when $1.44$ mole of C reacts $= \dfrac{{1.44}}{3} = 0.48$
Hence, the total number of moles of product formed will be $0.48$ .
Therefore, option (A) is correct.

Note:
The limiting reagent in a chemical reaction is that reactant that is consumed completely after the reaction is completed. The amount of product that is formed is limited by this reagent, as the reaction cannot proceed further without the limiting reagent.