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3 years ago

Consider the reaction below.

Chemistry

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

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%

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For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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