Question

Consider the reaction of 19.0 g of iron as follows.
4 Fe + 3 02 → 2 Fe2O3
(a) How many moles of iron react?
mol
(b) How many moles of oxygen must react?
mol
(c) How many moles of product form?
mol
(d) What mass of product forms?
9

Answer 1

Taking into account the reaction stoichiometry, you obtain:

• 0.3403 moles of iron react.
• 0.2551 moles of oxygen must react.
• 0.17 moles of Fe₂O₃ are formed.
• 27.6 grams of Fe₂O₃ are formed.

Reaction stoichiometry

In first place, the balanced reaction is:

4 Fe + 3 O₂  →  2 Fe₂O₃

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:

• Fe: 4 moles
• O₂: 3 moles
• Fe₂O₃: 2 moles

The molar mass of the compounds is:

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

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

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

Moles of Fe required

You can apply the following rule of three: If by definition of molar mass 55.85 grams are contained in 1 mole, 19 grams are contained in how many moles?

moles of Fe= (19 grams×1 mole) ÷ 55.85 grams

$moles of Fe=\frac{19 gramsx1 mole}{55.85 grams}$

moles of Fe= 0.3402 moles

Finally, 0.3403 moles of iron react.

Moles of O₂ required

The following rule of three can be applied: If by reaction stoichiometry 223.4 grams of Fe react with 3 moles of O₂, 19 grams of Fe react with how many moles of O₂?

moles of O₂= (19 grams×3 mole) ÷ 223.4 grams

moles of O₂= 0.2551 moles

Finally, 0.2551 moles of oxygen must react.

Moles of Fe₂O₃ formed

The following rule of three can be applied: if by reaction stoichiometry 223.4 grams of Fe form 2 mole of Fe₂O₃, 19 grams of Fe form how many moles of Fe₂O₃?

moles of Fe₂O₃=(19 grams×2 mole) ÷ 223.4 grams

moles of Fe₂O₃= 0.17 moles

Finally, 0.17 moles of Fe₂O₃ are formed.

Mass of Fe₂O₃ formed

The following rule of three can be applied: if by reaction stoichiometry 223.4 grams of Fe form 319.4 grams of Fe₂O₃, 19 grams of Fe form how many moles of Fe₂O₃?

msss of Fe₂O₃=(19 grams×319.4 grams) ÷ 223.4 grams

mass of Fe₂O₃= 27.6 grams

Finally, 27.6 grams of Fe₂O₃ are formed.

Summary

In summary, you obtain:

• 0.3403 moles of iron react.
• 0.2551 moles of oxygen must react.
• 0.17 moles of Fe₂O₃ are formed.
• 27.6 grams of Fe₂O₃ are formed.

Learn more about the reaction stoichiometry:

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