Consider the reaction below. 4Fes2 + 11O2 —> 2Fe2O3 + 8SO2
If the reaction produces 2.00 moles of Fe2O3, when 7.55 moles of O2 react, what is the percent yield of the reaction?

3. 68.55 g

4. 145.99%

Explanation:

3. Determination of the actual yield of NO.

4NH₃ + 5O₂ —> 4NO + 6H₂O

Next, we shall determine the mass of NH₃ that reacted and the NO that reacted from the balanced equation. This is illustrated below:

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mass of NH₃ from the balanced equation = 4 × 17 = 68 g

Molar mass of NO = 14 + 16

= 30 g

Mass of NO from the balanced equation = 4 × 30 = 120 g

Summary:

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO.

Next, we shall determine the theoretical yield of NO. This can be obtained as follow:

From the balanced equation above,

68 g of NH₃ reacted to produce 120 g of NO.

Therefore, 45.7 g of NH₃ will react to produce = (45.7 × 120)/68 = 80.65 g of NO.

Thus, the theoretical yield of NO is 80.65 g

Finally, we shall determine the actual yield of NO. This can be obtained as follow:

Percentage yield of NO = 85%

Theoretical yield of NO = 80.65 g

Actual yield of NO =?

Percentage yield = Actual yield /Theoretical yield × 100

85% = Actual yield / 80.65

85/100 = Actual yield / 80.65

0.85 = Actual yield / 80.65

Cross multiply

Actual yield of NO = 0.85 × 80.65

Actual yield of NO = 68.55 g

4. Determination of the percentage yield.

4FeS₂ + 11O₂ —> 2Fe₂O₃ + 8SO₂

From the balanced equation above,

11 moles of O₂ reacted to produce 2 moles of Fe₂O₃.

Next, we shall determine the theoretical yield of Fe₂O₃. This can be obtained as follow:

From the balanced equation above,

11 moles of O₂ reacted to produce 2 moles of Fe₂O₃.

Therefore, 7.55 moles of O₂ will react to produce = (7.55 × 2)/11 = 1.37 moles of Fe₂O₃.

Thus, the theoretical yield of Fe₂O₃ is 1.37 moles.

Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of Fe₂O₃ = 2 moles

Theoretical yield of Fe₂O₃ = 1.37 moles

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 2 / 1.37 ×100

Percentage yield = 200 / 1.37

Percentage yield = 145.99%

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.