Question

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

Answer 1

There will be made 8.7 grams of Al2O3

Step-by-step 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

Answer 2

8.66 g of Al₂O₃ will be produced

Step-by-step 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