A student working in the laboratory produces 6.81 grams of calcium oxide, CaO, from 20.7 grams of calcium carbonate, CaCO3, according to the reaction CaCO3 → CaO + CO2. Calculate the theoretical yield and
the percent yield for the reaction. Show and/or explain your work.

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%

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong