What volume of gas will be released when 500 g of marble (Ca-Co3), containing 10% impurities

What volume of gas will be released when 500 g of marble (Ca-Co3), containing 10% impurities, interacts with hydrochloric acid

1. Let’s find a mass of pure marble without admixtures.

100% – 10% = 90% (pure marble).

500 g – 100%,

X – 90%,

X = (500 g × 90%): 100 = 450 g.

2.Let’s find the amount of marble substance (calcium carbonate CaCO3) by the formula:

n = m: M.

M (CaCO3) = 40 + 12 + 48 = 100 g / mol.

n = 450 g: 100 g / mol = 4.5 mol.

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

CaCO3 + 2HCl = CaCl2 + CO2 + H2O.

According to the reaction equation, 1 mole of carbon dioxide accounts for 1 mole of calcium carbonate. Substances are in quantitative ratios 1: 1. The amount of marble and carbon dioxide will be the same.

n (CO2) = 4.5 mol.

3.V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 4.5 mol × 22.4 L / mol = 100.8 L.

Answer: 100.8 liters.