Calculate the mass of limestone containing 10% of impurities if, during its firing, calcium oxide weighing 112 g was obtained?

1) The main component of limestone is calcium carbonate – CaCO3. He underwent firing, that is, thermal decomposition took place. Write the equation of a chemical reaction:
CaCO3 = tC => CaO + CO2 ↑;

2) In the problem, we need the molar masses of CaCO3 and CaO:
Mr (CaCO3) = Ar (Ca) + Ar (C) + Ar (O) * 3 = 40 + 12 + 16 * 3 = 100 g / mol;
Mr (CaO) = Ar (Ca) + Ar (O) = 40 + 16 = 56 g / mol;

3) Units of measurement correspond to the SI system (%, g);

4) The known substance is calcium oxide – CaO. Find the amount of CaO:
n (CaO) = m (CaO) / Mr (CaO) = 112/56 = 2 mol;

3) Given the coefficients in front of CaCO3 and CaO in the reaction equation (1 and 1), the amount of CaCO3 will be equal to the amount of CaO:
n (CaCO3) = n (CaO) = 2 mol;

4) Using the formula, we find the mass of the CaCO3 substance:
m (CaCO3) = n (CaCO3) * Mr (CaCO3) = 2 * 100 = 200 g;

5) According to the condition, the content of impurities in the limestone sample is given. The rest is CaCO3. We find the mass fraction of CaCO3 contained in limestone:
ω (CaCO3) = 100% – ω (approx.) = 100% – 10% = 90%;

6) Knowing the mass of CaCO3 and its mass fraction, we find the mass of all limestone:
m (extract) = m (CaCO3) * 100% / ω (CaCO3) = 200 * 100% / 90% = 222 g.

Answer: The mass of limestone is 222 g.