Aluminum and oxygen react according to the following equation: 4Al (s) 3O2 (g) - -> 2Al2O3 (s) What mass of Al2O3, in grams, can be made by reacting 4.6 g Al with excess oxygen

8.66 g of Al₂O₃ will be produced

Explanation:

4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)

This is the reaction.

Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.

4.6 g / 26.98 g/mol = 0.170 moles

Ratio is 4:2.

4 moles of aluminum can produce 2 moles of Al₂O₃

0.170 moles of Al, may produce (0.170.2) / 4 = 0.085 moles

Let's convert the moles of Al₂O₃ to mass.

0.085 mol. 101.96 g/mol = 8.66 g

There will be made 8.7 grams of Al2O3

Explanation:

Step 1: Data given

Mass of Al = 4.6 grams

Molar mass Al = 26.98 g/mol

Oxygen is in excess

Molar mass of Al2O3 = 101.96 g/mol

Step 2: The balanced equation

4Al (s) + 3O2 (g) → 2Al2O3 (s)

Step 3: Calculate moles Al

Moles Al = Mass Al / molar mass Al

Moles Al = 4.6 grams / 26.98 g/mol

Moles Al = 0.1705 moles

Step 4: Calculate moles Al2O3

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

For 0.1705 moles Al we'll have 0.1705 / 2 = 0.08525 moles Al2O3

Step 5: Calculate mass Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 0.08525 moles * 101.96 g/mol

Mass Al2O3 = 8.7 grams

There will be made 8.7 grams of Al2O3