With the action of an excess of nitric acid on 50 g of calcium carbonate, 20 g of CO2 were obtained

With the action of an excess of nitric acid on 50 g of calcium carbonate, 20 g of CO2 were obtained, calculate the mass fraction of the yield from the theoretically possible.

Let’s find the amount of substance СaСO3.

n = m: M.

M (СaСO3) = 100 g / mol.

n = 50 g: 100 g / mol = 0.5 mol.

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

СaСO3 + 2 HNO3 = Ca (NO3) 2 + CO2 + H2O

According to the reaction equation, there is 1 mol of CO2 for 1 mol of СaСO3. The substances are in quantitative ratios of 1: 1.

n (СaСO3) = n (СO2) = 0.5 mol.

Let’s find the mass of CO2.

M (CO2) = 44 g / mol.

m = n × M.

m = 44 g / mol × 0.5 mol = 22 g (in theory).

22 g – 100%,

20 g – x%,

X = (20 × 100%): 22 g = 90.91%.

Answer: 90.91%.