How many Al2O3 can be created from 16 Al and excess O2? 

How many Al2O3 can be created from 18 O2 and excess Al?

How many Al are required to create 22 Al2O3? 

How many O2 are required to create 12 Al2O3?

How many Al are required to fully react with 30 O2? 

We are dealing with the chemical reaction for the formation of aluminum oxide (Al₂O₃) from aluminum (Al) and oxygen (O₂), which is given by the balanced chemical equation:

\[
4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3
\]

Let's address each part of the question using stoichiometry based on this equation.

### 1. How many Al₂O₃ can be created from 16 Al and excess O₂?

From the balanced equation:
- 4 moles of Al produce 2 moles of Al₂O₃.

Using a ratio:

\[
\frac{2 \text{ moles Al}_2\text{O}_3}{4 \text{ moles Al}} = \frac{x \text{ moles Al}_2\text{O}_3}{16 \text{ moles Al}}
\]

Solving for \(x\):

\[
x = \frac{2}{4} \times 16 = 8 \text{ moles Al}_2\text{O}_3
\]

So, **8 moles of Al₂O₃** can be created from 16 moles of Al.

### 2. How many Al₂O₃ can be created from 18 O₂ and excess Al?

From the balanced equation:
- 3 moles of O₂ produce 2 moles of Al₂O₃.

Using a ratio:

\[
\frac{2 \text{ moles Al}_2\text{O}_3}{3 \text{ moles O}_2} = \frac{x \text{ moles Al}_2\text{O}_3}{18 \text{ moles O}_2}
\]

Solving for \(x\):

\[
x = \frac{2}{3} \times 18 = 12 \text{ moles Al}_2\text{O}_3
\]

So, **12 moles of Al₂O₃** can be created from 18 moles of O₂.

### 3. How many Al are required to create 22 Al₂O₃?

From the balanced equation:
- 2 moles of Al₂O₃ require 4 moles of Al.

Using a ratio:

\[
\frac{4 \text{ moles Al}}{2 \text{ moles Al}_2\text{O}_3} = \frac{x \text{ moles Al}}{22 \text{ moles Al}_2\text{O}_3}
\]

Solving for \(x\):

\[
x = \frac{4}{2} \times 22 = 44 \text{ moles Al}
\]

So, **44 moles of Al** are required to create 22 moles of Al₂O₃.

### 4. How many O₂ are required to create 12 Al₂O₃?

From the balanced equation:
- 2 moles of Al₂O₃ require 3 moles of O₂.

Using a ratio:

\[
\frac{3 \text{ moles O}_2}{2 \text{ moles Al}_2\text{O}_3} = \frac{x \text{ moles O}_2}{12 \text{ moles Al}_2\text{O}_3}
\]

Solving for \(x\):

\[
x = \frac{3}{2} \times 12 = 18 \text{ moles O}_2
\]

So, **18 moles of O₂** are required to create 12 moles of Al₂O₃.

### 5. How many Al are required to fully react with 30 O₂?

From the balanced equation:
- 3 moles of O₂ react with 4 moles of Al.

Using a ratio:

\[
\frac{4 \text{ moles Al}}{3 \text{ moles O}_2} = \frac{x \text{ moles Al}}{30 \text{ moles O}_2}
\]

Solving for \(x\):

\[
x = \frac{4}{3} \times 30 = 40 \text{ moles Al}
\]

So, **40 moles of Al** are required to fully react with 30 moles of O₂.

---

Would you like more details or have any questions?

### Related Questions:
1. How many moles of Al₂O₃ can be produced from 24 moles of Al and 15 moles of O₂?
2. If 10 moles of Al₂O₃ are formed, how many moles of Al and O₂ are used?
3. How many grams of Al₂O₃ can be produced from 54 grams of Al and excess O₂?
4. What is the limiting reactant if 10 moles of Al and 10 moles of O₂ are used?
5. How many grams of O₂ are required to react with 50 grams of Al?

### Tip:
Always ensure the chemical equation is balanced before performing stoichiometry calculations!

How many Al2O3 can be created from 16 Al and excess O2?
How many Al2O3 can be created from 18 O2 and excess Al?
How many Al are required to create 22 Al2O3?
How many O2 are required to create 12 Al2O3?
How many Al are required to fully react with 30 O2?

Learn how to calculate the amount of Al2O3 produced in various reactions involving Al and O2 using stoichiometric ratios. This problem covers key chemistry concepts, including stoichiometry, chemical equations, and proportions. Suitable for students in Grades 10-12.

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