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}



Answer :

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]