When interacting sodium carbonate weighing 10.6 g with sulfuric acid weighing 19.6 g, find the mass of sodium sulfate salt.

m (Na2CO3) = 10.6 g.

m (H2SO4) = 19.6 g.

Determine the mass of sodium sulfate salt. We write down the solution.

First, we write down the equation of reactions.

Na2CO3 + H2SO4 = Na2SO4 + H2O + CO2

Find the amount of sodium carbonate and sulfuric acid substance, use the following formula.

n = m / M.

M (Na2CO3) = 23 * 2 + 12 + 3 * 16 = 106 g / mol.

M (H2SO4) = 1 * 2 + 32 + 16 * 4 = 98 g / mol.

n (Na2CO3) = 10.6 / 106 = 0.1 mol.

n (H2SO4) = 19.6 / 98 = 0.2 mol.

Sulfuric acid in excess, which means we count on sodium carbonate.

0.1 mol Na2CO3 – x mol Na2SO4

1 mol – 1 mol

X = 0.1 * 1: 1 = 0.1 mol Na2SO4.

M (Na2SO4) = 23 * 2 + 32 + 16 * 4 = 142 g / mol.

m = 142 * 0.1 = 14.2 g.

Answer: the mass of sodium sulfate is 14.2 g.