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
