Question

For the balanced equation shown below, if the reaction of 20.7 grams of CaCO₃ produces 6.81 grams of CaO, what is the theoretical and percent yield? (Ca = 40.08 amu, C = 12.01 amu, O = 16.00 amu)

CaCO₃ → CaO + CO₂

Answer 1

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = 20.7 g

mol of CaCO3 = (mass)/(molar mass)

= 20.7/100.09

= 0.2068 mol

From balanced chemical reaction, we see that

when 1 mol of CaCO3 reacts, 1 mol of CaO is formed

mol of CaO formed = moles of CaCO3

= 0.2068 mol

Molar mass of CaO = 1*MM(Ca) + 1*MM(O)

= 1*40.08 + 1*16.0

= 56.08 g/mol

mass of CaO = number of mol * molar mass

= 0.2068*56.08

= 11.6 g

% yield = actual mass*100/theoretical mass

= 6.81*100/11.6

= 58.7 %

11.6 g

58.7 %