The answer is #83.3%#.
From the balanced chemical equation
#CaCN_2 + 3H_2O -> CaCO_3 + 2NH_3#
we can see that we have a #1:2# mole ratio between #CaCN_2# and #NH_3#; that is, for every mole of #CaCN_2# used in the reaction, 2 moles of #NH_3# will be formed.
In order to determine the percent yield, we must determine what the limiting reagent is. We know that #CaCN_2# and #NH_3#'s molar masses are #80g/(mol)# and #17g/(mol)#, respectively, therefore
#n_(CaCN_2) = m_(CaCN_2)/(molar mass) = (77g)/(80.0g/(mol)) = 0.96# moles and
#n_(NH_3) = m_(NH_3)/(molarmass) = (27.1g)/(17g/(mol)) = 1.6# moles
According to our mole-to-mole ratio, #0.96# moles of #CaCN_2# should have produced #2 * 0.96 = 1.92# moles of #NH_3#, but instead produced #1.6# moles. This means that not all #CaCN_2# reacted <=> water is the limiting reagent.
So, the percent yield, defined as the actual yield divided by the theoretical yield and multiplied by 100%, is
#%yield = (actual)/(the o re tic) * 100% = (1.6 mol es)/(1.92 mol es) * 100% = 83.3%#
