the balanced equation for the neutralisation reaction is as follows 2HNO₃ + Ba(OH)₂ ---> Ba(NO₃)₂ + 2H₂O stoichiometry of HNO₃ to Ba(OH)₂ is 2:1 number of Ba(OH)₂ mol present - 0.108 mol 1 mol of Ba(OH)₂ neutralises 2 mol of Ba(OH)₂ therefore 0.108 mol of Ba(OH)₂ neutralises - 2 x 0.108 mol = 0.216 mol of HNO₃
