# Determine the mass of sodium sulfate that is formed by the interaction of solutions containing 12 g

**Determine the mass of sodium sulfate that is formed by the interaction of solutions containing 12 g of sodium hydroxide and 19.6 g of sulfuric acid.**

1. Write down the reaction equation.

2NaOH + H2SO4 = Na2SO4 + 2H2O.

Find the molar mass of sodium hydroxide and sulfuric acid.

M (NaOH) = 23 + 16 + 1 = 40 g / mol.

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

We find the mass of sulfuric acid.

12 g NaOH – X g H2SO4

2 × 40 g / mol NaOH – 98 g / mol H2SO4

X = 12 × 98 / (2 × 40) = 14.7 g.

This means that sulfuric acid should be 14.7. Consequently, sulfuric acid is in excess. We calculate sodium sulfate by sodium hydroxide.

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

12 g NaOH – x g Na2SO4.

2 × 40 g / mol NaOH – 142 g / mol Na2SO4

X = 12 × 142 / (2 × 40) = 21.3 g.

Answer: m (Na2SO4) = 21.3 g.