In 200 g of a solution with a mass fraction of acetic acid of 30% was placed 26.5 g

| July 29, 2021 | education

In 200 g of a solution with a mass fraction of acetic acid of 30% was placed 26.5 g of sodium carbonate. Determine the mass fraction of sodium acetate in the resulting solution.

The reaction of obtaining sodium acetate.
2CH3COOH + Na2CO3 = 2CH3COONa + H2O + CO2.
A mass of acetic acid that has reacted with sodium carbonate.
m (CH3COOH) = W (CH3COOH) • m (solution) / 100% = 30% • 200 g / 100% = 60 g.
The total mass of the solution.
m (solution) = m (solution) ref. + m (Na2CO3) = 200 g + 26.5 g = 226.5 g.
Determine the number of moles of reagents.
n (CH3COOH) = m / Mr = 60 g / 2 * 60 g / mol = 0.5 mol.
n (Na2CO3) = 26.5 g / 106 g / mol = 0.25 mol.
The calculation of the amount of mol of acetate is carried out according to the lack.
n (CH3COONa) = 2 • n (Na2CO3) = 2 • 0.25 mol = 0.5 mol.
The mass of the obtained sodium acetate.
m (CH3COONa) = n • Mr = 0.5 mol • 82 g / mol = 41 g.
Mass fraction.
W (CH3COONa) = m (CH3COONa) / m (solution) • 100% = 41 g / 226.5 g • 100% = 18%.



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