How much water should be taken to slake lime obtained from limestone weighing 200 kg?
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;