Question

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxide . iron ( s ) + oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide ? 21.6 grams What is the percent yield for this reaction ? 85 %

Answer 1

The question relates to calculating the theoretical and percent yield of iron(II) oxide from a chemical reaction between iron and oxygen. The theoretical yield has been stated as 21.6 grams, and the percent yield as 85%. However, without a balanced chemical equation and proper stoichiometric calculations provided, the theoretical yield cannot be confirmed, and the actual percent yield calculation with the given numbers does not match the stated 85%.

Step-by-step explanation:

The question is asking for the theoretical and percent yield of iron(II) oxide when iron reacts with excess oxygen. In stoichiometry, the theoretical yield is the maximum amount of product that can be produced from a given amount of reactants, while the percent yield is a comparison of the actual yield to the theoretical yield, calculated as (actual yield/theoretical yield) × 100%.

To calculate the theoretical yield of iron(II) oxide, we would need to use the balanced equation for the reaction between iron and oxygen. However, the information provided is incomplete for this calculation. The student has provided the value of 21.6 grams as the theoretical yield without the necessary calculations or a balanced equation presented, and a percent yield of 85%. To determine the percent yield, you would divide the actual yield of iron(II) oxide (5.17 grams) by the theoretical yield (21.6 grams) and multiply by 100%.

However, without a balanced chemical equation and complete stoichiometric calculations, we cannot confirm the provided theoretical yield. If the provided theoretical yield of 21.6 grams is correct, then the percent yield calculation would be (5.17 grams / 21.6 grams) × 100%, giving an actual percent yield of 23.94%, not 85%.

Answer 2

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

Step-by-step 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}=(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 =
`(2)/(2)* 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}}* 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}=(5.17g)/(5.53g)* 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 %