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.

Answer:

  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.