Answer to Question #136079 in General Chemistry for Sera

Question

During the manufacture of cement for the construction industry limestone, CaCO3

(s), is decomposed to quicklime, CaO, by vigorous heating. What is the percentage yield and percent difference of the reaction when 13.1 tonne (1 Mg = 1 tonne=106g) of quicklime are obtained from the treatment of 24.5 tonne of limestone?

CaCO3(s) → CaO(s) + CO2(g)

Answer 1

$$n(CaCO_3) = m/M ={24.5*10^6g \over 100g/mol} = 24.5*10^4 mol$$

$$n(CaCO_3) = n(CaO) = 24.5*10^4mol = 2.45*10^5mol$$

From data:

$$n(CaO) = {13.1*10^6g \over 56g/mol} = 2.34*10^5mol$$

Then:

$$\eta = {2.34*10^5 \over 2.45*10^5} = 0.955 = 95.5\%$$