How much water should be taken to slake lime obtained from limestone weighing 200 kg? Water is required in threefold excess. Mass fraction of calcium carbonate in limestone is 90%. Take the density of water equal to 1 kg / l.
The reaction of calcining limestone and obtaining slaked lime is described by the following equations:
CaCO3 = CaO + CO2 ↑;
CaO + H2O = Ca (OH) 2;
When decomposing 1 mol of limestone, 1 mol of calcium oxide and 1 mol of carbon monoxide is obtained. From this amount, it is possible to obtain 1 mol of calcium hydroxide.
Let’s determine the chemical amount of calcium carbonate. To do this, divide its weight by its molar mass.
M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol; N CaCO3 = 200,000 x 0.9 / 100 = 1,800 mol;
To obtain slaked lime, you need to take 1,800 x 3 = 5,400 moles of water.
Let’s calculate its weight.
For this purpose, we multiply its weight by its molar weight.
M H2O = 2 + 16 = 18 grams / mol; m H2O = 5 400 x 18 = 97 200 grams = 97.2 kg = 97.2 liters;
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