# Find the mass of the salt, the yield of which was 95%, if it is formed by the interaction

**Find the mass of the salt, the yield of which was 95%, if it is formed by the interaction of 200 grams of 14% sodium hydroxide solution with orthophosphate acid.**

Find the mass of NaOH in solution.

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (NaOH) = (200 g × 14%): 100% = 28 g.

Find the amount of NaOH.

M (NaOH) = 40 g / mol.

n = m: M.

n = 28 g: 40 g / mol = 0.7 mol.

Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

3NaOH + H3PO4 = Na3PO4 + 3H2O.

According to the reaction equation, there is 1 mol of Na3PO4 per 3 mol of NaOH. The substances are in quantitative ratios of 3: 1.

The amount of Na3PO4 is 3 times less than the amount of NaOH.

n (Na3PO4) = 1/3 n (NaOH) = 0.7: 3 = 0.23 mol.

Let’s find the mass of Na3PO4.

m = n × M,

M (Na3PO4) = 164 g / mol.

m = 0.23 mol × 164 g / mol = 37.72 g (theory).

37.72 – 100%,

X – 95%,

X = (37.72 × 95%): 100% = 35.83 g.

Answer: 35.82 g.