# Find the mass of salt that is formed by draining 200g of 30% sodium hydroxide solution

**Find the mass of salt that is formed by draining 200g of 30% sodium hydroxide solution with a solution containing 49g of sulfuric acid**

Find the mass of NaOH in solution.

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

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

m (NaOH) = (200 g × 30%): 100% = 60 g.

Find the amount of NaOH.

M (NaOH) = 40 g / mol.

n = m: M.

n = 60 g: 40 g / mol = 1.5 mol.

Let’s find the amount of the substance H2SO4.

M (H2SO4) = 98 g / mol.

n = m: M.

n = 49 g: 98 g / mol = 0.5 mol (deficiency).

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

2 NaOH + H2SO4 = Na2SO4 + 2H2O.

According to the reaction equation, there is 1 mol of Na3PO4 for 1 mol of H2SO4. Substances are in quantitative ratios 1: 1.

The amount of substance Na3PO4 and H2SO4 are equal.

n (Na2SO4) = n (H2SO4) = 0.5 mol.

Find the mass of Na2SO4.

m = n × M,

M (Na2SO4) = 142 g / mol.

m = 0.5 mol × 142 g / mol = 71 g.

Answer: 71 g.