Question

What is the total number of moles of aluminum oxide that can be formed when 54 grams of aluminum reacts completely with oxygen?

Answer 1

4Al + 3O₂ → 2Al₂O₃

m(Al)=54 g
M(Al)=27 g/mol

n(Al₂O₃)=n(Al)/2

n(Al)=m(Al)/M(Al)

n(Al₂O₃)=m(Al)/{2M(Al)}

n(Al₂O₃)=54/{2*27}= 1 mol

Answer 2

Answer:

1.0 mol

Explanation:

Let's consider the synthesis of aluminum oxide.

2 Al + 1.5 O₂ → Al₂O₃

The molar mass of Al is 27 g/mol. The moles represented by 54 g of Al are:

54 g × (1 mol/ 27 g) = 2.0 mol

The molar ratio of Al to Al₂O₃ is 2:1. The moles of Al₂O₃ are:

2.0 mol Al × (1 mol Al₂O₃/2 mol Al) = 1.0 mol

1.0 mol of Al₂O₃ is formed.