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

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 = ?

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.

To determine the percentage yield of the transformation, we need to compare the actual yield of calcium oxide (CaO) with the theoretical yield of CaO.

First, we need to calculate the molar mass of CaO and CO2:

- Molar mass of CaO (Ca = 40, O = 16):
= 40 + 16
= 56 g/mol

- Molar mass of CO2 (C = 12, O = 16):
= 12 + (16 x 2)
= 44 g/mol

Next, we can calculate the theoretical yield of CaO by using stoichiometry and the balanced equation:

1 mole of CaCO3 produces 1 mole of CaO

Given:
- Mass of limestone sample = 71.42 g
- Mass of CaO obtained = 13 g

We can now calculate the moles of CaCO3 reacted:

Moles of CaO = Mass of CaO / Molar mass of CaO
= 13 g / 56 g/mol
= 0.232 moles

Since 1 mole of CaCO3 produces 1 mole of CaO, the moles of CaCO3 reacted is also 0.232.

Now we need to calculate the theoretical yield of CaCO3:

Theoretical yield of CaCO3 = Moles of CaCO3 x Molar mass of CaCO3
= 0.232 moles x (40 + 12 + (16 x 3)) g/mol
= 10.4 g

Finally, we can calculate the percentage yield using the actual and theoretical yields:

Percentage Yield = (Actual Yield / Theoretical Yield) x 100
= (13 g / 10.4 g) x 100
= 125%

Therefore, the percentage yield of the transformation is 125%.