The neutralization reaction with sulfuric acid entered 300 grams of 5% sodium hydroxide solution

| September 12, 2021 | education

The neutralization reaction with sulfuric acid entered 300 grams of 5% sodium hydroxide solution. Calculate the mass of the salt formed. You can write in detail what we take from where.

Find the mass of sodium hydroxide in the solution.

The mass fraction of a substance is calculated by the formula:

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%,

m (substance) = (300 g × 5%): 100% = 15 g.

Find the amount of sodium hydroxide substance by the formula:

n = m: M.

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

n = 15 g: 40 g / mol = 0.375 mol.

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

2NaOH + H2SO4 = Na2SO4 + 2H2 O.

According to the reaction equation, for 2 mol of sodium hydroxide there is 1 mol of the salt obtained – sodium sulfate. Substances are in quantitative ratios of 2: 1.

The amount of the salt substance will be 2 times less than the amount of the sodium hydroxide substance.

n (Na2SO4) = 2n (NaOH) = 0.375: 2 = 0.1875 mol.

Find the mass of sodium sulfate.

m = n × M.

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

m = 142 g / mol × 0.1875 mol = 26.625 g.

Answer: 26.625 g.



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