Calculate the volume of carbon monoxide (IV) and the mass of calcium oxide that can be obtained

Calculate the volume of carbon monoxide (IV) and the mass of calcium oxide that can be obtained with the complete decomposition of 500 g of limestone? which contains 10 %% impurities.

Limestone formula CaCO3 (calcium carbonate). Its molar mass is 100 g / mol. Of the available 500 g, 10% are impurities that do not react. Calcium carbonate mass = 500 g * 0.9 = 450 g. Let us find the amount of calcium carbonate substance = 450 g / 100 g / mol = 4.5 mol.

The thermal decomposition of calcium carbonate.

CaCO3 = CaO + CO2.

In the equation, all substances are assigned 1 to 1. Therefore, from 4.5 moles of the starting substance, 4.5 moles of each of the products are formed.

Find the volume of nitrogen dioxide = 4.5 * 22.4 liters = 100.8 liters.

Find the mass of calcium oxide = 4.5 * 56 = 252 g.

Answer: the volume of CO2 = 100.8 liters, the mass of CaO = 252 g.

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