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]
