From 1.25 kg of limestone containing 20% impurities, during its decomposition, you can get how much carbon dioxide

Limestone is essentially calcium carbonate. When heated strongly, it decomposes into carbon dioxide and calcium oxide, also called quicklime. The equation for this reaction can be written as:

CaCO3 = CaO + CO2

To solve our problem, we first calculate how much pure calcium carbonate was contained in a sample weighing 1.2 kg. And then we calculate the mass and volume of carbon dioxide. But first, let’s convert the mass from kilograms to grams for correct calculations.

1.2 kg. = 1200 g.

From the condition we know that the content of impurities in limestone is 20%. Means pure carbonate 100% – 20% = 80%

m = 1200 * 80/100 = 960 g of pure calcium carbonate

Molar mass of calcium carbonate = 100 g / mol;

Molar mass of carbon dioxide = 44 g / mol

Molar volume of carbon dioxide = 22.4 l / mol

We make a proportion for finding both the mass and the volume of the gas:

960 g of calcium carbonate corresponds to X g of carbon dioxide or Y l. as

100 g / mol of calcium carbonate corresponds to 44 g / mol of carbon dioxide or 22.4 l / mol

X = 960 * 44/100 = 422.4 g of carbon dioxide

Y = 960 * 22.4 / 100 = 215.04 l. carbon dioxide

Answer: 422.4 g or 215.04 liters of CO2