How many moles of hno3 are present if 0.108 mol of ba(oh)2 was needed to neutralize the acid solution? express your answer with the appropriate units?

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

5.g i think hope it helps u not