Sodium hydroxide, taken in the required amount, was treated with a solution containing 252 g of nitric acid.

Sodium hydroxide, taken in the required amount, was treated with a solution containing 252 g of nitric acid. Calculate the mass of the salt obtained if the yield is practically 90% of the theoretical.

1. Let’s write the reaction equation:

NaOH + HNO3 = NaNO3 + H2O.

2. Find the amount of nitric acid:

n (HNO3) = m (HNO3) / M (HNO3) = 252 g / 63 g / mol = 4 mol.

3. According to the reaction equation, we find the theoretical amount of sodium nitrate:

ntheor (NaNO3) = n (NaOH) = 4 mol.

4. Find the practical amount of salt, and then its mass:

nprak (NaNO3) = ntheor (NaNO3) * η (NaNO3) / 100% = 4 mol * 90% / 100% = 3.6 mol.

m (NaNO3) = nprak (NaNO3) * M (NaNO3) = 3.6 mol * 85 g / mol = 306 g.

Answer: m (NaNO3) = 306 g.