Question

In the reaction 2A+4B---->3C+4D, when 5 mols of A reacts with 6 moles of B, Then (i) which one is the limiting reagent (ii)Calculate the amount of "C" formed.

Answer 1

2A + 4B => 3C + 4D
1) 2 mols of A requires 4 mols of B for the reaction
Hence, for 5 mols of A, the mols of B required
                                              = 5 mols of A x 4 mols of B/2 mols of A
But as we have only 6 mols of B, ∴ B is the limiting reagent.

2) Amount of C is determined by B
4 mols of B gives 3 mols of C
Hence, 6 mols of B = 6 mol of B x 3 mol of C/ 4 mol of B
                                = 4.5 mol of C 

Answer 2

2A + 4B —> 3C + 4D

Here, 2 moles of A reacts with 4 moles of B. Therefore,

5 moles of A reacts with 4/2x5 =10 moles of B.

But,

we have 6 moles of ‘B’

(i) It means ‘B’ is limiting reactant

(ii) 4 moles of ‘B’ gives 3 moles of C.

Hence, 6 moles of ‘B’ gives

3/4 X6 = 18/4=9/2== 4.5 moles of ‘C’