Question

When 50.5 g iron(III) oxide reacts with carbon monoxide, 32.2 g iron is produced. What is the percent yield of the reaction?

Answer 1

Taking into account definition of percent yield, the percent yield for the reaction is 91.17%.

Reaction stoichiometry

In first place, the balanced reaction is:

Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• Fe₂O₃: 1 mole
• CO: 3 moles
• Fe: 2 moles
• CO₂: 3 moles

The molar mass of the compounds is:

• Fe₂O₃: 159.7 g/mole
• CO: 28 g/mole
• Fe: 55.85 g/mole
• CO₂: 44 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• Fe₂O₃: 1 mole ×159.7 g/mole= 159.7 grams
• CO: 3 moles ×28 g/mole= 84 grams
• Fe: 2 moles ×55.85 g/mole= 111.7 grams
• CO₂: 3 moles ×44 g/mole= 132 grams

Mass of Fe formed

The following rule of three can be applied: if by reaction stoichiometry 159.7 grams of Fe₂O₃ form 111.7 grams of Fe, 50.5 grams of Fe₂O₃ form how much mass of Fe?

$mass of Fe=\frac{50.5 grams of Fe_{2} O_{3} x111.7 grams of Fe}{159.7 grams of Fe_{2} O_{3}}$

mass of Fe= 35.32 grams

Then, 35.32 grams of Fe can be produced from 50.5 g iron(III) oxide.

Percent yield

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

$percent yield=\frac{actual yield}{theorical yield}x100$

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Percent yield for the reaction in this case

In this case, you know:

• actual yield= 32.2 grams
• theorical yield= 35.32 grams

Replacing in the definition of percent yields:

$percent yield=\frac{32.2grams}{35.32grams}x100$

Solving:

percent yield= 91.17%

Finally, the percent yield for the reaction is 91.17%.

Learn more about

the reaction stoichiometry:

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percent yield:

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