Is 100 ml of a 97% acetic acid solution of 1.04 g / ml density sufficient to dissolve 100 g

| January 10, 2021 | education

Is 100 ml of a 97% acetic acid solution of 1.04 g / ml density sufficient to dissolve 100 g of calcium carbonates? What is the mass of soda with this?

Let’s write the reaction equation:
2СН3СООН + CaCO3 = Ca (СН3СОО) 2 + H2O + CO2
Let’s determine the amount of substances – participants in the reaction.
n (CaCO3) = m (CaCO3) / M (CaCO3) = 100/100 = 1 mol, where m (CaCO3) is the mass of calcium carbonate, M (CaCO3) is the molar mass of calcium carbonate.
n (CH3COOH) = m (CH3COOH) / M (CH3COOH), where M (CH3COOH) is the molar mass of acetic acid, m (CH3COOH) is the mass of acetic acid.
We do not know the mass of acetic acid, we will find it, knowing the density p (CH3COOH), the volume of the solution V (CH3COOH), and the mass fraction w (CH3COOH). The product of volume and density is the mass of the acetic acid solution.
m (CH3COOH) = V (CH3COOH) * p (CH3COOH) * w (CH3COOH) / 100% = 100 * 1.04 * 97/100% = 100.88 g.
n (CH3COOH) = 100.88 / 60 = 1.68 mol
According to the reaction equation, 2 mol of acetic acid is required for 1 mol of calcium carbonate, and we got 1.68 mol. This means that acetic acid is in short supply, and the calculation of the amount of the formed calcium acetate salt must be carried out using acetic acid.
Then the amount of calcium carbonate reacted will be n (CaCO3) n = n (CH3COOH) / 2 = 1.68 / 2 = 0.84 mol. The same amount of mol is formed by calcium acetate n (ac) = 0.84
Calcium acetate mass m (ac) = n (ac) * M (ac) = 0.84 * 158 = 132.72 g
M (ac) – molar mass of calcium acetate



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