Mass of water produced : 0.146 g
Further explanation
Given
33.2 mL of 0.245 M lithium hydroxide
Required
mass of water
Solution
Reaction
HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)
mol LiOH :
= M x V
= 0.245 x 33.2 ml
= 8.134 mmol
From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol
mass H₂O :
= mol x MW
= 8.134 x 10⁻³ mol x 18 g/mol
= 0.146 g
