Calculate the mass of sodium sulfate formed by neutralizing sodium hydroxide with a mass of 12 g with sulfuric acid

When sodium hydroxide is neutralized with sulfuric acid, the average sodium sulfate salt and water are formed. Reaction at n.u. proceeds actively to the end. The reaction equation can be written as:

2 NaOH + H2SO4 = Na2SO4 + H2O

Molar mass of substances:

M (NaOH) = 40 g / mol;

M (Na2SO4) = 142 g / mol.

Let’s make a proportion according to the reaction equation and calculate the mass of the salt.

12 g NaOH corresponds to X g Na2SO4, as

2 * 40 g / mol NaOH corresponds to 142 g / mol Na2SO4.

X = 12 * 142 / (2 * 40) = 21.3 g of sodium sulfate.

Answer: the mass of sodium sulfate is 21.3 g.