A mixture of sodium carbonate and sodium bicarbonate completely reacted with 300 grams of 14% acetic acid

| September 23, 2021 | education

A mixture of sodium carbonate and sodium bicarbonate completely reacted with 300 grams of 14% acetic acid solution. this produced 11.2 liters of gas (normal conditions). determine the mass fraction of sodium carbonate in the original mixture.

1.Let’s find the mass of CH3COOH in solution.

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

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

m (substance) = (300 g × 14%): 100% = 42 g.

1.Let’s find the amount of CH3COOH substance.

M (CH3COOH) = 60 g / mol.

n = 42 g: 60 g / mol = 0.7 mol.

Let’s find the quantitative ratios of substances.

2СН3СООН + Na2CO3 = 2СН3СОО Na + CO2 + H2O.

For 2 mol of CH3COOH there is 1 mol of Na2CO3. The substances are in quantitative ratios of 2: 1. The amount of Na2CO3 is 2 times less than the amount of CH3COOH.

n (Na2CO3) = ½ n (CH3COOH) = 0.7: 2 = 0.35 mol.

Let’s find the mass of Na2CO3.

M (Na2CO3) = 106 g / mol.

m = nM.

m = 0.35 mol × 106 g / mol = 37.1 g.

n (CO2) = V: Vn

n (CO2) = 11.2 L: 22.4 L / mol = 0.5 mol.

CH3COOH + NaHCO3 = CH3COO Na + CO2 + H2O.

There is 1 mole of NaHCO3 per mole of CO2.

Substances are in quantitative ratios 1: 1.

The amount of substance CO2 and NaHCO3 are equal.

n (CO2) = n (NaHCO3) = 0.5 mol.

Let’s find the mass of NaHCO3.

M (NaHCO3) = 84 g / mol.

m = nM.

m = 0.5 mol × 84 g / mol = 42 g.

Let’s find the mass of the mixture.

m (mixture) = m (NaHCO3) + m (Na2CO3).

m = 42 g + 37.1 g = 79.1 g.

w = m (Na2CO3): m (mixture) × 100%

w (Na2CO3) = (37.1 g: 79.1 g) × 100% = 47%

Answer: 47%.



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