The limiting reagent is HCl.
The reaction's balanced chemical equation is
#Ca(OH)_(2(aq)) + 2HCl(aq) -> CaCl_2(aq) + 2H_2O_((l))#
We can see that we have a #1:2# mole ratio
between #Ca(OH)_2# and #HCl#; that is, every mole of #Ca(OH)_2# used in the reaction needs 2 moles of
#HCl#. We can determine the number of moles of #HCl# by using its
molarity, #C = n/V#
#n_(HCl) = C * V = 0.500M * 75.0 * 10^-3 L= 0.038# moles
We can determine the number of #Ca(OH)_2# moles to be (#Ca(OH)_2#'s molar mass is #74g/(mol e)#)
#n_(Ca(OH)_2) = m/(molarmass) = (15.0g)/(74g/(mol e)) = 0.20# moles:
Taking into consideration the mole-to-mole ration, #0.20# moles of #Ca(OH)_2# would have required # 2* 0.20 = 0.40# moles of #HCl#; we can clearly see that the number of #HCl# moles available is almost 10 times smaller, #0.038#.
Therefore, the limiting reagent is #HCl#.
As a side note, notice that since #HCl# is a strong acid, which means it dissociates completely into #H^+# and #Cl^-#, and since #CaCl_2# is a soluble salt which dissociates into #Ca^(2+)# and #Cl^-# anions, the reaction's complete ionic equation would be
#Ca(OH)_(2(aq)) + 2H^(+)(aq) + 2Cl^(-)(aq) -> Ca^(2+)(aq) + 2Cl^(-)(aq) + 2H_2O_((l))#
The net ionic equation would be (after removing the spectator ions)
#Ca(OH)_(2(aq)) + 2H^(+)(aq) -> Ca^(2+)(aq) + 2H_2O_((l))#
