CaCO3(s)-> CaO(s) + CO2(g)
Convert 71.42 g CaCO3 to mols. mols = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO.
Now convert mols CaO to grams. g CaO = mols CaO x molar mass CaO. This is the theoretical yield (TY). The actual yield (AY) in the problem is 13.0 grams CaO.
%yield = (AY/TY)*100 = ?
A sample of calcario and other components of the soil was intensely warm. Under these conditions, the calcario undergoes decomposition in calcium oxide and
carbon dioxide.
CaCO3 (s)-> CaO (s) + CO2 (g)
The analysis of 71, 42 g of a sample of limestone supplied 13 g of calcium oxide (CaO) and carbon dioxide (CO2), after high temperature heating. Determine the percentage yield of the transformation.
data: C = 12, Ca = 40, O = 16
3 answers
thanks @DrBOb222 :)
Assuming the 71.42gms of limestone is pure CaCO3 then,
(71.42g CaCO3/100gms/mole)=0.714mole CaCO3.
From CaCO3 => CaO + CO2,the theoretical yield is a 1:1 rxn ratio. Therefore, from 0.714 mole CaCO3, the reaction theoretically delivers 0.714mole CaO and 0.714mole CO2. Therefore, the theoretical gram yield of CaO = (0.714mole CaO)(56g/mole)=39.984g CaO and the theoretical yield of CO2 = (0.714mole CO2)(44g/mole) = 31.416g CO2. From this and 13gms of each product detected, %Yield CaO = (13/39.984)100% = 32.51 wt% CaO and the %Yield CO2 = (13/31.416)100% = 41.38 wt% CO2.
(71.42g CaCO3/100gms/mole)=0.714mole CaCO3.
From CaCO3 => CaO + CO2,the theoretical yield is a 1:1 rxn ratio. Therefore, from 0.714 mole CaCO3, the reaction theoretically delivers 0.714mole CaO and 0.714mole CO2. Therefore, the theoretical gram yield of CaO = (0.714mole CaO)(56g/mole)=39.984g CaO and the theoretical yield of CO2 = (0.714mole CO2)(44g/mole) = 31.416g CO2. From this and 13gms of each product detected, %Yield CaO = (13/39.984)100% = 32.51 wt% CaO and the %Yield CO2 = (13/31.416)100% = 41.38 wt% CO2.