Answer:
A. Theoretical yield of CaO is 11.59 g
B. Percentage yield of CaO = 58.76%
Explanation:
The following data were obtained from the question:
Mass of CaCO₃ = 20.7 g
Actual yield of CaO = 6.81 g
Theoretical yield of CaO =?
Percentage yield of CaO =?
The equation for the reaction is given below:
CaCO₃ —> CaO + CO₂
Next, we shall determine the mass of CaCO₃ that decomposed and the mass of CaO produced from the balanced equation. This can be obtained as follow:
Molar mass of CaCO₃ = 40 + 12 + (3×16)
= 40 + 12 + 48
= 100 g/mol
Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g
Molar mass of CaO = 40 + 16 = 56 g/mol
Mass of CaO from the balanced equation = 1 × 56 = 56 g
SUMMARY:
From the balanced equation above,
100 g of CaCO₃ decomposed to produce 56 g of CaO.
A. Determination of the theoretical yield of CaO.
From the balanced equation above,
100 g of CaCO₃ decomposed to produce 56 g of CaO.
Therefore, 20.7 g of CaCO₃ will decompose to produce =
(20.7 × 56)/100 = 11.59 g of CaO.
Thus, the theoretical yield of CaO is 11.59 g
B. Determination of the percentage yield.
Actual yield of CaO = 6.81 g
Theoretical yield of CaO = 11.59 g
Percentage yield of CaO =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield = 6.81/11.59 × 100
Percentage yield of CaO = 58.76%
