(a) Equation one: NH_3 +HCl \to NH_4Cl
Equation two: NaOH +HCl \to NaCl+H_2O
(b) to find the moles of excess HCl, we need to find the moles of NaOH present in the 23.30 mL of 0.116 mol L−1 sodium hydroxide solution-->
c=\frac{n}{v}\therefore 0.116 X 23.3x10^-3=2.7028x10^{-3} $Moles of NaOH...