Question

The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction $2A+4B\to 3C+4D,$ when 5 moles of A react with 6 moles of B, then

(i) Which is the limiting reagent ?

(ii) Calculate the amount of C formed ?

Answer 1

(i) $2A+4B⟶3C+4D$

According to the given reaction, 2 moles of A react with 4 moles of B.

Hence, 5 moles of A will react with 10 moles of $B(\frac{5\times 4}{2}=10moles)$

It indicates that reactant B is limiting reagent as it will consume first in the reaction because we have only 6 moles of B.

(ii) $2A+4B⟶3C+4D$

According to the given reaction, 2 moles of A react with 4 moles of B.

Hence, 5 moles of A will react with 10 moles of $B(\frac{5\times 4}{2}=10moles)$

Limiting reagent decide the amount of product produced.

According to the reaction.

4 moles of B produces 3 moles of C

$\therefore 6$ moles of B will produce $\frac{3\times 6}{4}=4.5$ moles of C.

Note Limiting reagent limits the amount of product formed because it is present in lesser amount and gets consumed first.