Under the action of 146 g of a hydrochloric acid solution with a mass fraction of 30% chlorine on calcium

| August 6, 2021 | education

Under the action of 146 g of a hydrochloric acid solution with a mass fraction of 30% chlorine on calcium carbonate with a mass of 25 g, a gas with a mass of 10 g was obtained. What is the mass fraction of the reaction product yield?

Find the mass of HCl in solution.

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

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

m (HCl) = (146 g × 30%): 100% = 43.8 g.

Find the amount of HCl.

М (HCl) = 36.5 g / mol.

n = m: M.

n (HCl) = 43.8 g: 36.5 g / mol = 1.2 mol (excess).

Let’s find the amount of substance СaСO3.

n = m: M.

M (СaСO3) = 100 g / mol.

n = 25 g: 100 g / mol = 0.25 mol.

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

СaСO3 + 2HCl = CaCl2 + CO2 + H2O

According to the reaction equation, 1 mol of СaСO3 accounts for 1 mol of СO2. Substances are in quantitative ratios 1: 1.

n (СaСO3) = n (СO2) = 0.25 mol.

Let’s find the mass of CO2.

M (CO2) = 44 g / mol.

m = n × M.

m = 44 g / mol × 0.25 mol = 11 g (according to theory).

11 g – 100%,

10 g – x%,

X = (10 × 100%): 11 g = 90.91%.

Answer: 90.91%.



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