As a result of the neutralization reaction, 14.2 g of sodium sulfate was formed. using the law of conservation

As a result of the neutralization reaction, 14.2 g of sodium sulfate was formed. using the law of conservation of mass, calculate the masses of the reacting acids, bases, and released water.

To solve the problem, you need to draw up an equation:
1. 2NaOH + H2SO4 = Na2SO4 + 2H2O – ion exchange, obtained sodium sulfate;
1 mol 1 mol;
2. Calculation of the molar masses of substances:
M (NaOH) = 39.9 g / mol;
M (H2SO4) = 98 g / mol;
M (Na2SO4) = 141.8 g / mol.
3. Determine the amount of salt, if its mass is known:
Y = m / M = 14.2 / 141.8 = 0.1 mol;
Y (H2SO4) = 0.1 mol since the amount of these substances according to the equation is 1 mol.
4. Proportion:
0.1 mol (Na2SO4) – X mol (NaOH);
-1 mol -2 mol hence, X mol (NaOH) = 0.1 * 2/1 = 0.2 mol.
5. Find the mass of the starting materials:
m (NaOH) = Y * M = 0.2 * 39.9 = 7.98 g;
m (H2SO4) = Y * M = 0.1 * 98 = 9.8 g.
Answer: the process will require sodium hydroxide weighing 7.98 g, sulfuric acid – 9.8 g.