Determine the volume and amount of substance (mol) of carbon monoxide 4

Determine the volume and amount of substance (mol) of carbon monoxide 4 that can be obtained by decomposition of 0.6 kg of limestone containing 5% impurities.

1. Let’s write down the reaction of limestone decomposition:

CaCO3 = CaO + CO2 ↑;

2.Let’s find the mass of pure calcium carbonate in limestone:

m (CaCO3) = (1 – w (impurities)) * m (limestone);

m (CaCO3) = (1 – 0.05) * 600 = 570 g;

3.Calculate the chemical amount of calcium carbonate and carbon monoxide (4):

n (CaCO3) = m (CaCO3): M (CaCO3);

M (CaCO3) = 40 + 12 + 48 = 100 g / mol;

n (CaCO3) = 570: 100 = 5.7 mol;

n (CO2) = n (CaCO3) = 5.7 mol;

4.determine the volume of carbon dioxide:

V (CO2) = n (CO2) * Vm = 5.7 * 22.4 = 127.68 dm3.

Answer: 5.7 mol; 127.68 dm3.