Question

What mass (in g) of iron (iii) oxide is needed to produce 155 g of iron if this reaction is known to have a 68.9% yield

Answer 1

Taking into account the reaction stoichiometry, 321.63 grams of iron (III) oxide is needed to produce 155 g of iron if this reaction is known to have a 68.9% yield.

Reaction stoichiometry

In first place, the balanced reaction is:

2 Fe₂O₃ → 4 Fe + 3 O₂

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: 2 moles
• Fe: 4 moles
• O₂: 3 moles

The molar mass of the compounds is:

• Fe₂O₃: 159.7 g/mole
• Fe: 55.85 g/mole
• O₂: 32 g/mole

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

• Fe₂O₃: 2 moles ×159.7 g/mole= 319.4 grams
• Fe: 4 moles ×55.85 g/mole= 223.4 grams
• O₂: 3 moles ×32 g/mole= 96 grams

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%:

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

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.

Mass of iron (II) oxide needed

In this case, you know:

• percent yield= 68.9 %
• actual yield= 155 grams
• theorical yield= ?

Replacing in the definition of percent yields:

$68.9=\frac{155 grams}{theorical yield} x100$

Solving:

$\frac{68.9}{100} =\frac{155 grams}{theorical yield}$

$0.689 =\frac{155 grams}{theorical yield}$

0.689× theorical yield= 155 grams

theorical yield= 155 grams ÷0.689

theorical yield= 224.96 grams

The following rules of three can be applied: If by reaction stoichiometry 223.4 grams of Fe are produced from 319.4 grams of Fe₂O₃, 224.96 grams of Fe are produced from how much mass of Fe₂O₃?

$mass of Fe_{2} O_{3} =\frac{224.96 grams of Fex319.4 gramsof Fe_{2} O_{3}}{223.4 grams of Fe}$

mass of Fe₂O₃=  321.63 grams

Finally, 321.63 grams of iron (III) oxide is needed to produce 155 g of iron if this reaction is known to have a 68.9% yield.

Learn more about

the reaction stoichiometry:

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percent yield:

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