2al+ fe203 al203 +2fe in one process, 121 g of al are reacted with 601 g of fe203. calculate the mass (in grams) of al203 formed, and determine the amount of excess reagent left at the end of the reaction.

229 g Al₂O₃; 243 g Fe₂O₃

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.  

M_r:      26.98    159.69    101.96

              2Al   +   Fe₂O₃ ⟶ Al₂O₃ + 2Fe

Mass/g:  121          601

Step 2. Calculate the moles of each reactant

Moles of Al         = 121 × 1/26.98

Moles of Al         = 4.485 mol Al

Moles of Fe₂O₃  = 601× 1/159.69

Moles of Fe₂O₃  = 3.764 mol Fe₂O₃

Step 3. Identify the limiting reactant  

Calculate the moles of Al₂O₃ we can obtain from each reactant.  

From Al

The molar ratio is 1 mol Al₂O₃:2 mol Al

Moles of Al₂O₃ = 4.485 × 1/2

Moles of Al₂O₃ = 2.242 mol Al₂O₃

From Fe₂O₃:

The molar ratio is 1 mol Al₂O₃:1 mol Fe₂O₃

Moles of Al₂O₃ = 3.764 × 1/1

Moles of Al₂O₃ = 3.764 mol Al₂O₃

The limiting reactant is Al because it gives the smaller amount of Al₂O₃.

The excess reactant is Fe₂O₃.

Step 4. Calculate the mass of Al₂O₃ formed

Mass of Al₂O₃ = 2.242 × 101.96

Mass of Al₂O₃ = 229 g Al₂O₃

Step 5. Calculate the moles of Fe₂O₃ reacted

The molar ratio is 1 mol Fe₂O₃:2 mol Al:

Moles of Fe₂O₃ = 4.485 × ½

Moles of Fe₂O₃ = 2.242 mol Fe₂O₃

Step 6. Calculate the moles of Fe₂O₃ remaining

Moles remaining = original moles – moles used

Moles remaining = 3.764 – 2.242

Moles remaining = 1.521 mol Fe₂O₃

Step 7. Calculate the mass of Fe₂O₃ remaining

Mass of Fe₂O₃ = 1.521 × 159.69/1

Mass of Fe₂O₃ = 243 g Fe₂O₃

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