What mass of calcium carbonate in the form of limestone must be taken to produce 150 kg of quicklime

What mass of calcium carbonate in the form of limestone must be taken to produce 150 kg of quicklime from it, if it is known that the practical yield of the product is 85% of the theoretical?

The limestone decomposition reaction is described by the following chemical reaction equation:
CaCO3 = CaO + CO2;
The decomposition of 1 mol of limestone synthesizes 1 mol of gaseous carbon dioxide.
This also synthesizes 1 mol of calcium oxide
Let’s calculate the chemical amount of a substance in 150 kg of quicklime.
M CaO = 40 + 16 = 56 grams / mol;
N CaO = 150,000/56 = 2 679 mol;
Taking into account the reaction yield of 85%, to obtain such an amount of substance, it is necessary to take 2 679 / 0.85 = 3 152 mol of lime.
Let’s calculate its weight:
M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;
m CaCO3 = 3 152 x 100 = 315.2 kg;