What volume (liters) of aqueous 0.325 M nitric acid is required to react completely with 3.68 g barium hydroxide?

Question

asked Aug 14, 2021 147k views

1 Answer

RSW · Answer 1 · 2021-08-20T16:54:16+0000

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) :

$\tt mol=(mass)/(MW)\\\\mol=(3.68)/(171,34 g/mol)\\\\mol=0.0215$

From the equation, mol ratio of HNO₃ : Ba(OH)₂ = 2 : 1, so mol HNO₃:

$\tt mol~HNO_3=(2)/(1)* 0.0215=0.043$

Molarity of HNO₃ = 0.325, then the volume of HNO₃ :

$\tt V=(n(mol))/(M(molarity))\\\\V=(0.043)/(0.325)\\\\V=0.132~L$