What volume of carbon monoxide (IV) can be obtained by reacting nitric acid with 20 g of calcium

What volume of carbon monoxide (IV) can be obtained by reacting nitric acid with 20 g of calcium carbonate containing 3% impurities?

1. Let us write down the equations of the flowing interaction:

CaCO3 + 2HNO3 = Ca (NO3) 2 + H2O + CO2 ↑;

2.Calculate the mass of calcium carbonate:

m (CaCO3) = w (CaCO3) * m (sample);

w (CaCO3) = 1 – w (impurities) = 1 – 0.03 = 0.97;

m (CaCO3) = 0.97 * 20 = 19.4 g;

3. find the chemical amount of carbonate:

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

M (CaCO3) = 40 + 12 + 3 * 16 = 100 g / mol;

n (CaCO3) = 19.4: 100 = 0.194 mol;

4.determine the amount of carbon dioxide:

n (CO2) = n (CaCO3) = 0.194 mol;

5.Calculate the volume of released gas:

V (CO2) = n (CO2) * Vm = 0.194 * 22.4 = 4.35 liters.

Answer: 4.35 liters.