How many grams of [tex]$Al_2O_3$[/tex] can form from [tex]33.8 \, g[/tex] of [tex]Al[/tex]?

Given the reaction:
[tex]\[ 4Al(s) + 3O_2(g) \longrightarrow 2Al_2O_3(s) \][/tex]

1. Show the strategy for solving this problem.

[tex]\[ \text{grams } Al \longrightarrow \text{moles } Al \longrightarrow \text{moles } Al_2O_3 \longrightarrow \text{grams } Al_2O_3 \][/tex]

\begin{tabular}{|lll|}
\hline
Answer Bank & \\
\hline
moles [tex]$Al_2O_3$[/tex] & moles [tex]$O_2$[/tex] & grams [tex]$Al_2O_3$[/tex] \\
grams [tex]$Al$[/tex] & grams [tex]$O_2$[/tex] & moles [tex]$Al$[/tex] \\
\hline
\end{tabular}

To find the amount of $\text{Al}_2\text{O}_3$ formed from a given mass of $\text{Al}$, we can follow a step-by-step strategy using stoichiometry principles based on the balanced chemical equation:

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

### Step-by-Step Strategy:

1. Convert grams of $ \text{Al} $ to moles of $ \text{Al} $:
- Use the molar mass of Aluminium ($ \text{Al} $) to convert the given mass to moles.
- [tex]\[ \text{grams} \, \text{Al} \longrightarrow \text{moles} \, \text{Al} \][/tex]

2. Convert moles of $ \text{Al} $ to moles of $ \text{Al}_2\text{O}_3 $:
- Use the stoichiometric ratio from the balanced chemical equation to find the moles of $ \text{Al}_2\text{O}_3 $.
- [tex]\[ \text{moles} \, \text{Al} \longrightarrow \text{moles} \, \text{Al}_2\text{O}_3 \][/tex]

3. Convert moles of $ \text{Al}_2\text{O}_3 $ to grams of $ \text{Al}_2\text{O}_3 $:
- Use the molar mass of Aluminium Oxide ($ \text{Al}_2\text{O}_3 $) to convert the computed moles to grams.
- [tex]\[ \text{moles} \, \text{Al}_2\text{O}_3 \longrightarrow \text{grams} \, \text{Al}_2\text{O}_3 \][/tex]

Now, let's plug in the values and walk through each step:

### Detailed Solution:

1. Given data:
- Mass of $ \text{Al} $ = 33.8 grams
- Molar mass of $ \text{Al} $ = 26.98 g/mol
- Molar mass of $ \text{Al}_2\text{O}_3 $ = 101.96 g/mol

2. Convert grams of $ \text{Al} $ to moles of $ \text{Al} $:
- [tex]\[ \text{moles} \, \text{Al} = \frac{\text{grams} \, \text{Al}}{\text{molar mass} \, \text{Al}} = \frac{33.8 \, \text{g}}{26.98 \, \text{g/mol}} \approx 1.253 \, \text{moles} \, \text{Al} \][/tex]

3. Using the balanced equation, convert moles of $ \text{Al} $ to moles of $ \text{Al}_2\text{O}_3 $:
- According to the equation, 4 moles of $ \text{Al} $ produce 2 moles of $ \text{Al}_2\text{O}_3 $. Thus,
- [tex]\[ \text{moles} \, \text{Al}_2\text{O}_3 = \left(\frac{\text{moles} \, \text{Al}}{4}\right) \times 2 = \left(\frac{1.253}{4}\right) \times 2 \approx 0.626 \, \text{moles} \, \text{Al}_2\text{O}_3 \][/tex]

4. Convert moles of $ \text{Al}_2\text{O}_3 $ to grams of $ \text{Al}_2\text{O}_3 $:
- [tex]\[ \text{grams} \, \text{Al}_2\text{O}_3 = \text{moles} \, \text{Al}_2\text{O}_3 \times \text{molar mass} \, \text{Al}_2\text{O}_3 = 0.626 \, \text{moles} \times 101.96 \, \text{g/mol} \approx 63.867 \, \text{grams} \][/tex]

Thus, from 33.8 grams of $\text{Al}$, approximately 63.867 grams of $\text{Al}_2\text{O}_3$ can be formed.

So the final answer is:
[tex]\[ \boxed{63.867 \, \text{grams}} \][/tex]