How much CO2 will be released during the decomposition of 500 g of limestone containing 10% impurities?
April 28, 2021 | education
| The decomposition reaction of calcium carbonate occurs in accordance with the following chemical reaction equation:
CaCO3 = CaO + CO2;
From 1 mole of calcium carbonate, 1 mole of calcium oxide and 1 mole of carbon dioxide are formed.
The weight of pure calcium carbonate is 500 x 0.9 = 450 grams.
Let’s determine the chemical amount of a substance in 450 grams of calcium carbonate.
M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;
N CaCO3 = 450/100 = 4.5 mol;
Let’s calculate the volume of 4.5 mol of carbon dioxide.
One mole of gas under normal conditions fills a volume of 22.4 liters.
The volume of carbon dioxide will be equal to:
V CO2 = 4.5 x 22.4 = 100.8 liters;
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