• # question_answer 79)   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 : (a) which is the limiting reagent? (b) calculate the amount of C formed.

The given reaction is : $\underset{2mol}{\mathop{2A}}\,+\underset{4mol}{\mathop{4B}}\,\to \underset{3mol}{\mathop{3C}}\,+\underset{4mol}{\mathop{4D}}\,$ Given moles of A and B are 5 and 6 moles respectively. Case I:   Let reactant (A) is completely consumed. 2 mol A $\equiv$ 3 mol C $\therefore$  $5molA\equiv \frac{3}{2}\times 5molC\equiv 7.5\,molC$ Case II: Let reactant (B) is completely consumed. 4 mol 5 $\equiv$ 3 mol C $\therefore$  $6\text{ }mol\text{ }B\equiv \frac{3}{4}\times 6mol\text{ }C\text{ }\equiv \text{ }4.5\text{ }mol\text{ }C$ Since (B) on complete consumption gives least amount of product (C) hence (B) will be limiting reagent and the amount of (C) formed will be 4.S mol.