Question

Use the reaction,  8 Al + 3 Fe₃O₄ = 4 Al₂O₃ + 9 Fe

(a) How many grams of iron are produced by the reaction of 225.0 grams of Al and 225.0 grams of Fe₃O₄?

(b) How many grams of  Al₂O₃ are also produced in the reaction of (a)?
(c) Which reactant is in excess and how many grams remain after reaction?

Answer 1

Concepts used to solve:

• To calculate molar mass of a compound, add the atomic masses of each component atom.
• Equ1: No of moles =
• To find how much substance is produced or consumed, always compare the number of moles from the balanced chemical reaction.
• Limiting reagent = reagent present in lesser quantity than needed for both reactants to fully react. Excess reagent = reagent present in excess, and some of it remains unreacted.
• Yield of product is calculated considering that the limiting reagent has reacted completely.

Balanced reaction :

8 Al + 3 Fe₃O₄ = 4 Al₂O₃ + 9 Fe

Calculating moles of each reactant :

225.0 g of Al is taken. Molar mass of Al = 26.98 g/mol

Using the above formula, no. of moles of Al = = 8.34 moles

225.0 g of Fe₃O₄ is taken. Molar mass of Fe₃O₄ = 231.53 g/mol

Using the above formula, no. of moles of Fe₃O₄ = = 0.972 moles

Finding limiting reagent :

From the balanced equation, 8 mole Al reacts with 3 moles Fe₃O₄

Thus, 8.34 moles Al reacts with = 3.13 moles Fe₃O₄.

However, moles of Fe₃O₄ taken is much lesser than this. So, Fe₃O₄ is the limiting reagent, Al is the excess reactant.

(a)

From the balanced equation, 3 mole Fe₃O₄ gives 9 moles Fe

Thus, 0.972 moles Fe₃O₄ gives = 2.916 moles Fe.

Mass of Fe (molar mass = 55.845 g/mol) (using equ1) = number of moles x molar mass = 2.916 moles x 55.845 g/mol = 162.8 g

(b)

From the balanced equation, 3 mole Fe₃O₄ gives 4 moles Al₂O₃

Thus, 0.972 moles Fe₃O₄ gives = 1.296 moles Al₂O₃.

Mass of Al₂O₃ (molar mass = 102 g/mol) (using equ1) = number of moles x molar mass = 1.296 moles x 102 g/mol = 132.2 g

(c)

From the balanced equation, 3 mole Fe₃O₄ needs 8 moles Al

Thus, 0.972 moles Fe₃O₄ needs   = 2.592 moles Al

So, moles of Al remaining = Total taken ( found in the beginning)- moles reacted = 8.34 moles – 2.592 moles = 5.748 moles

Mass of Al (molar mass = 27 g/mol) (using equ1) = number of moles x molar mass = 5.748 moles x 102 g/mol = 155.2 g