In the first reaction 1 mole of A reacts with 2 moles of B to form product P1. Hence, B is the limiting reagent here, which will give 1 mole of P1. Now, in 2nd reaction 1 mole of P1 will require 0.75 mole of C. Then, 4 moles of P1 will require 4×0.75=3 moles of C. Hence, C will be completely exhausted.