Aluminum oxide forms when aluminum reacts with oxygen. 4Al (s) + 3O2 (g) → 2Al2O3 (s) A mixture of 82.49 g of aluminum (M. W. = 26.98 g/mol) and 117.65 g of oxygen (M. W. = 32.00 g/mol) is allowed to react. What mass of aluminum oxide (M. W. = 101.96 g/mol) can be formed?

Question

Chemistry

Garrett Dodson

2 October, 19:35

Aluminum oxide forms when aluminum reacts with oxygen. 4Al (s) + 3O2 (g) → 2Al2O3 (s) A mixture of 82.49 g of aluminum (M. W. = 26.98 g/mol) and 117.65 g of oxygen (M. W. = 32.00 g/mol) is allowed to react. What mass of aluminum oxide (M. W. = 101.96 g/mol) can be formed?

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Answers (2)

Know the Answer?

Browning · Answer 1

155.9 grams of aluminium oxide can be formed

Explanation:

Step 1: Data given

Mass of aluminium = 82.49 grams

Molar mass of aluminium = 26.98 g/mol

Mass of oxygen = 117.65 grams

Molar mass of oxygen = 32.0 g/mol

Molar mass of aluminium oxide = 101.96 g/mol

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate moles

Moles = mass / molar mass

Moles aluminium = 82.49 grams / 26.98 g/mol

Moles aluminium = 3.057 moles

Moles oxygen = 117.65 grams / 32.0 g/mol

Moles oxygen = 3.677 moles

Step 4: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Aluminium is the limiting reactant. It will completely be consumed (3.057 moles). Oxygen is in excess. There will react 3/4 * 3.057 = 2.293 moles

There will remain 3.677 - 2.293 = 1.384 moles

Step 5: Calculate moles aluminium oxide

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

For 3.057 moles Al we'll have 3.057/2 = 1.529 moles

Step 6: Calculate mass aluminium oxide

Mass aluminium oxide = 1.529 moles * 101.96 g/mol

Mass aluminium oxide = 155.9 grams

155.9 grams of aluminium oxide can be formed

Gabrielle Monroe · Answer 2

155.4 g of Al₂O₃

Explanation:

The reaction is:

4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

To determine the mass of aluminum oxide that is formed we need to know the limiting reagent. Let's calculate the moles of each by the molar mass

Mass / molar mass = Moles

82.49 g / 26.98 g/mol = 3.05 moles of Al

117.65 g / 32 g/mol = 3.67 moles of oxygen

Let's try the oxygen. Ratio is 3:4.

3 moles of O₂ need 4 moles of Al to react

Therefore 3.67 moles of O₂ will react with (3.67. 4) / 3 = 4.90 moles

We only have 3.05 moles of Al, so the Al is the limiting reactant

Now, we work with stoichiometry

4 moles of Al can produce 2 moles of Al₂O₃

3.05 moles of Al will produce (3.05.2) / 4 = 1.52 moles of Al₂O₃

We convert the moles to mass: 1.52 mol. 101.96 g / 1mol = 155.4 g