# 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.