For the decomposition of 1 mol of calcium carbonate, 157 kJ of heat must be spent.

For the decomposition of 1 mol of calcium carbonate, 157 kJ of heat must be spent. Determine the amount of heat required to decompose 650 kg of calcium carbonate.

Decomposition of calcium carbonate:

CaCO3 = CaO + CO2.

We find the molar mass of calcium carbonate: M (Ca) = 40 g / mol, M (C) = 12 g / mol, M (O) = 16 g / mol.

M (CaCO3) = 40 + 12+ 16×3 = 100 g / mol.

We find the amount of substance in 650 kg of calcium carbonate:

650 kg = 650,000 g.
650,000 (g): 100 (g / mol) = 6500 mol.

For the decomposition of 1 mol of a substance, 157 kJ of heat is needed, for the decomposition of 6500 mol, respectively:

157 x 6500 = 1,020,500 kJ = 1,020.5 MJ of heat.