Question

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

Answer 1

The reaction is shown below.

$$2\mathrm{A}+4\mathrm{B}⟶3\mathrm{C}+4\mathrm{D}$$

2 moles of A requires 4 moles of B. So, the number of moles of B required for 5 moles of A is calculated as,

Moles of B = 5 mol of A × ${\displaystyle \frac{4 \text{mol of B}}{2 \text{mol of A}}}$ = 10 mol of B

So, the required number of moles of B is 10 mol but only 6 moles of B are given in the question.