Determine the mass of the salt formed by the action of a sulfuric acid solution on 20 g of sodium hydroxide.

Let’s execute the solution:
1. In accordance with the condition of the problem, we compose the equation:
m = 20 g. X g. -?
H2SO4 + 2NaOH = Na2SO4 + 2H2O – ion exchange, sodium sulfate is released;
2. Calculations:
M (NaOH) = 22.9 +17 = 39.9 g / mol;
M (Na2SO4) = 22.9 * 2 + 32 + 64 = 141.8 g / mol;
Y (NaOH) = m / M = 20 / 39.9 = 0.5 mol.
3. Proportion:
0.5 mol (NaOH) – X mol (Na2SO4);
– 2 mol – 1 mol from here, X mol (Na2SO4) = 0.5 * 1/2 = 0.25 mol.
4. Find the mass of the product:
m (Na2SO4) = Y * M = 0.25 * 141.8 = 35.45 g.
Answer: the mass of sodium sulfate salt is 35.45 g.