# Sodium in an amount of 0.5 mol was dissolved in water.

**Sodium in an amount of 0.5 mol was dissolved in water. Calculate the mass of sulfuric acid required to neutralize the resulting sodium hydroxide.**

Let’s compose the reaction equation, find the quantitative ratios of substances.

2 Na + 2H2O = 2 NaOH + H2.

According to the reaction equation, 2 mol of sodium accounts for 2 mol of NaOH. Substances are in quantitative ratios 1: 1.

n (Na) = n (NaOH) = 0.5 mol.

2NaOH + H2SO4 = Na2SO4 + 2H2O.

According to the reaction equation, there is a mole of H2SO4 for 2 mol of NaOH. The substances are in quantitative ratios of 2: 1. The amount of sulfuric acid substance is 2 times less than the amount of NaOH substance.

n (H2SO4) = ½ n (NaOH) = 0.5: 2 = 0.25 mol.

Let’s find the mass of Н2SO4.

m = n × M,

M (H2SO4) = 98 g / mol.

m = 0.25 mol × 98 g / mol = 24.5 g.

Answer: 24.5 g.