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

Chemistry

1 answer:

Tamiku [17]3 years ago

8 0

Volume (liters) of aqueous 0.325 M nitric acid is 0.132 L

<h3>Further explanation</h3>

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) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{3.68}{171,34 g/mol}\\\\mol=0.0215$

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

$\tt mol~HNO_3=\dfrac{2}{1}\times 0.0215=0.043$

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

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

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