The limiting reagent is CaC_2.
First, start with the balanced chemical equation
CaC_2 + 2H_2O -> C_2H_2 + Ca(OH)_2
Notice that we have a 1:2 mole ratio between CaC_2 and H_2O; that is, every mole of the former used requires 2 moles of the latter.
Since the quantities of both reactants are given, and knowing that their molar masses are 62.0 g/(mol) (for CaC_2) and 18.0 g/(mol) (for H_2O), we can determine the number of moles from
n_(H_2O) = m_(H_2O)/(molarmass_(H_2O)) = (33.71 g)/(18 g/(mol)) = 1.87 moles
n_(CaC_2) = m_(CaC_2)/(molarmass_(CaC_2)) = (43.25 g)/(64.0 g/(mol)) = 0.68 moles
Notice that the number of moles of CaC_2 determined would require 2 * 0.68 = 1.36 moles of H_2O, less than what we have in the reaction. Therefore, H_2O is in excess, which means that CaC_2 is the limiting reagent.
