How many grams of Al2O3 are produced when 60.2 g of Al reacts? express your answer with the appropriate units.

113.1 g of Al₂O₃ are produced

Explanation:

We firstly determine the reaction:

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

If you see stoichiometry, we make react 4 moles of Al in order to produce 2 moles of Al₂O₃.

Let's convert the mass of Al, to moles and then, we make a rule of three:

60.2 g / 26.98 g/mol =2.23 moles

4 moles of Al can produce 2 moles of alumina

Therefore, 2.23 moles will produce (2.23 . 2) /4 = 1.11 moles of Al₂O₃

We convert the moles to mass: 1.11 mol . 101.96 g /1mol = 113.1 g

114 grams of Al2O3 will be produced

Explanation:

Step 1: Data given

Mass of Al = 60.2 grams

Molar mass of Al = 26.98 g/mol

Molar mass Al2O3 = 101.96 g/mol

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate moles Al

Moles Al = Mass Al / molar mass Al

Moles Al = 60.2 grams / 26.98 g/mol

Moles Al = 2.23 moles

Step 4: Calculate moles Al2O3

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

For 2.23 moles Al we'll have 2.23/ 2 = 1.115 moles Al2O3

Step 5: Calculate mass Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 1.115 moles * 101.96 g/mol

Mass Al2O3 = 113.7 grams (≈114 grams)

114 grams of Al2O3 will be produced