Volume (liters) of aqueous 0.325 M nitric acid is 0.132 L
Reaction
2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O
mass of Ba(OH)₂ = 3.68 g
mol Ba(OH)₂(MW=171,34 g/mol) :
[tex]\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{3.68}{171,34 g/mol}\\\\mol=0.0215[/tex]
From the equation, mol ratio of HNO₃ : Ba(OH)₂ = 2 : 1, so mol HNO₃:
[tex]\tt mol~HNO_3=\dfrac{2}{1}\times 0.0215=0.043[/tex]
Molarity of HNO₃ = 0.325, then the volume of HNO₃ :
[tex]\tt V=\dfrac{n(mol)}{M(molarity)}\\\\V=\dfrac{0.043}{0.325}\\\\V=0.132~L[/tex]
