# 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;