Upon neutralization of 294 g of sulfuric acid with sodium hydroxide, 400 g of sodium sulfate was formed

Upon neutralization of 294 g of sulfuric acid with sodium hydroxide, 400 g of sodium sulfate was formed. What is the mass fraction (in%) of the salt yield from the theoretically possible?

Let’s implement the solution:

In accordance with the condition of the problem, we write down the equation of the process:
H2SO4 + 2NaOH = Na2SO4 + 2H2O – ion exchange, sodium sulfate is released;

Calculations by formulas:
M (H2SO4) = 98 g / mol;

M (Na2SO4) = 141.8 g / mol;

Y (H2SO4) = m / M = 294/98 = 3 mol;

Y (Na2SO4) = 3 mol since the amount of substances is 1 mol.

Find the mass and yield of the product:
m (Na2SO4) = Y * M = 3 * 141.8 = 425.4 g (theoretical weight);

W = m (practical) / m (theoretical) * 100;

W = 400 / 425.4 * 100 = 94.03%

Answer: the yield of sodium sulfate was 94.03%