Calculate and note the mass fraction of etanoic acid in a solution weighing 150 g

| April 30, 2021 | education

Calculate and note the mass fraction of etanoic acid in a solution weighing 150 g, if a sodium hydroxide solution weighing 250 g with a mass fraction of alkali of 16% was spent on its neutralization.

1. Let’s write down the equality of chemical interaction:

CH3COOH + NaOH = CH3COONa + H2O.

2. Find the chemical number of sodium hydroxide:

m (NaOH) = m (solution) * ω (NaOH) / 100% = 250 g * 16% / 100% = 40 g.

n (NaOH) = m (NaOH) / M (NaOH) = 40 g / 40 g / mol = 1 mol.

3. Based on the equality of the chemical interaction, we find the chemical number and, as a result, the mass fraction of acetic acid:

n (CH3COOH) = n (NaOH) = 1 mol.

m (CH3COOH) = n (CH3COOH) * M (CH3COOH) = 1 mol * 60 g / mol = 60 g.

ω (CH3COOH) = m (CH3COOH) / m (solution) * 100% = 60 g / 150 g * 100% = 40%.

Answer: ω (CH3COOH) 40%.



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