If 16.9 g of CO2 was produced from the thermal decomposition of 41.83 g of CaCO3, what is the percentage yield of the reaction?

Question

CaCO3(s) > CaO(s) + CO2(g)

Answer 1

Calculate the theoretical yield

41.83 g of CaCO3 *(1 mole/100.08 g of CaCO3)= moles of CaCO3

SInce the mole ratios are equal, moles of CaCO3=CO2 moles of CO2

moles of CO2*(44.01 g of CO2/moles of CO2)= g of CO2 from the theoretical reaction.

(16.9 g of CO2 was produced/g of CO2 from theoretical rxn)*100= % yeild

Answer 2

Thanks!