Question

Given the balanced equation:
4Al (s) + 3O2 (g) --> 2Al2O3 (s)
What is the total number of moles of O2 (g) that must react completely with 8.0 moles of Al (s) in order to form Al2O3?

Answer 1

To find the total number of moles of O₂ that must react completely with 8.0 moles of Al, we can use the stoichiometry of the balanced chemical equation:

\[ 4 \text{Al (s)} + 3 \text{O}_2 (g) \rightarrow 2 \text{Al}_2\text{O}_3 (s) \]

From the balanced equation, we can see the mole ratio of aluminum (Al) to oxygen (O₂) is 4:3. This means for every 4 moles of aluminum, 3 moles of oxygen are required.

Now, we can set up a ratio to find the amount of O₂ needed for 8.0 moles of Al:

\[
\frac{3 \text{ moles O}_2}{4 \text{ moles Al}} = \frac{x \text{ moles O}_2}{8.0 \text{ moles Al}}
\]

We can cross-multiply to solve for \( x \):

\[
3 \text{ moles O}_2 \cdot 8.0 \text{ moles Al} = 4 \text{ moles Al} \cdot x \text{ moles O}_2
\]

This simplifies to:

\[
24 = 4x
\]

Now, divide both sides by 4:

\[
x = \frac{24}{4} = 6
\]

So, the total number of moles of O₂ that must react completely with 8.0 moles of Al to form Al₂O₃ is **6.0 moles of O₂**.