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 %

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:

      ....(1)

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

For the given chemical reaction:

By Stoichiometry of the reaction:

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

So, 0.0771 moles of iron will produce = 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:  

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

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:

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

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