Consider the given reaction and statements:

$2\mathrm{A} + 3\mathrm{B} \to \mathrm{C}$

Statement -1:  $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ moles of 'C' are always produced when 3 moles of 'A' & 4 moles of 'B' are                           added.

Statement -2 : 'B' is the liming reactant for the given data.

$2\mathrm{A} + 3\mathrm{B} \to \mathrm{C}$

Given information of the first statement that 3 moles of A and 4 moles of B produce $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ moles of C.

Explanation:

From the reaction

$2\mathrm{A} + 3\mathrm{B} \to 1\mathrm{C}$

It can conclude that, to prepare 1 mole of C; we require 3 moles of B and 2 moles of A. It can be said that for preparing x moles of C, 3x moles of B and 2x moles of A are required.

So, to prepare $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ moles of C , :

$\frac{4}{3}\times 3$ moles of B and $\frac{4}{3}\times 2$ moles of A

Hence, 4 moles of B and 2.6 moles of A require to produce $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ moles of C.

This means that first statement is false.

According to the question, there are 3 moles of A to produce $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ moles of 'C', but required moles is only 2.6 moles. Hence,  A is excess reagent and B is limiting Reagent.

So, Statement 1 is false and statement 2 is true.