What volume of carbon monoxide 4 can be obtained from the decomposition of 400 g of calcium carbonate containing 6 percent impurities?
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 400 x 0.94 = 376 grams.
Let’s determine the chemical amount of a substance in 376 grams of calcium carbonate.
M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;
N CaCO3 = 376/100 = 3.76 mol;
Let’s calculate the volume of 3.76 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 = 3.76 x 22.4 = 84.22 liters;
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