Calculate the mass of the salt formed by the interaction of 6.3 g of nitric acid with a solution containing 2 g of sodium hydroxide.

m (HNO3) = 6.3 g.
m (NaOH) = 2 g.
Let’s determine the mass of the resulting salt. We write down the solution.
First, we write down the equation of reactions.
HNO3 + NaOH = NaNO3 + H2O
Find the molar mass of nitric acid and sodium hydroxide.
M (HNO3) = 1 + 14 + 3 * 16 = 63 g / mol.
M (NaOH) = 23 + 16 + 1 = 40 g / mol.
Determine how much sodium hydroxide should be.
6.3 g HNO3 – x g NaOH
63 g / mol HNO3 – 40 g / mol NaOH
X = 6.3 * 40: 63 = 4 g.
This means that sodium hydroxide is in short supply. Find the mass of salt by sodium hydroxide.
M (NaNO3) = 23 + 14 + 16 * 3 = 85 g / mol.
2 g NaOH – x g NaNO3
40 g / mol NaOH – 85 g / mol NaNO3
X = 2 * 85: 40 = 4.25 g.
Answer: m (NaNO3) = 4.25 g.