Under the action of an excess of hydrochloric acid on calcium carbonate with a mass of 25 g, carbon monoxide (IV)

Under the action of an excess of hydrochloric acid on calcium carbonate with a mass of 25 g, carbon monoxide (IV) with a mass of 10 g was obtained. Determine the product yield.

Let’s find the amount of substance CaCO3.

n = m: M.

M (CaCO3) = 100 g / mol.

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

Let’s find the quantitative ratios of substances.

CaCO3 + 2HCl → CaCl2 + H2O + CO2 ↑.

For 1 mol of CaCO3, there is 1 mol of CO2.

Substances are in quantitative ratios 1: 1.

The amount of substance will be equal.

n (CO2) = n (CaCO3) = 0.25 mol.

Let’s find the mass of CO2.

m = n × M.

M (CO2) = 44 g / mol.

m = 44 g / mol × 0.25 mol = 11 g.

11 g should be obtained by calculations (theory), according to the condition of the problem, the output mass of carbon dioxide is 10.

Let’s find the mass of CO2 from the theoretically possible.

11 – 100%,

10 g – x%.

x = (100% × 10): 11% = 90.9%.

Answer: 90.9%.