Calculate the mass of water resulting from the interaction of 100g. nitric acid and lithium hydroxide.

Let’s write the reaction equation:
HNO3 + LiOH = LiNO3 + H2O
It can be seen from the reaction equation that:
ν (CH3COOH) = ν (NaOH)
m (HNO3) / M (HNO3) = m (H2O) / M (H2O)
m (H2O) = m (HNO3) * M (H2O) / M (HNO3)
Let’s define the molar masses of water and acid:
M (HNO3) = 1 + 14 + 16 * 3 = 63 g / mol
M (H2O) = 1 * 2 + 16 = 18 g / mol
Let’s calculate the mass of water:
m (H2O) = m (HNO3) * M (H2O) / M (HNO3) = 100 * 18/63 = 28.6 g.
Answer: the mass of water released as a result of acid neutralization is 28.6 g.