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

| September 30, 2021 | education

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.



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