Sure, let's solve this problem step by step to find the mass of aluminum oxide (Al₂O₃) that could form from 8.0 grams of oxygen (O₂) with an excess of aluminum (Al).
### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
[tex]$
4 \, \text{Al (s)} + 3 \, \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{Al}_2\text{O}_3 \, \text{(s)}
$[/tex]
### Step 2: Determine the molar masses
- Molar mass of [tex]\(\text{O}_2\)[/tex] is [tex]\(32.00 \, \text{g/mol}\)[/tex]
- Molar mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] is [tex]\(101.96 \, \text{g/mol}\)[/tex]
### Step 3: Convert the given mass of [tex]\(\text{O}_2\)[/tex] to moles
We are given 8.0 grams of [tex]\(\text{O}_2\)[/tex].
To find the number of moles of [tex]\(\text{O}_2\)[/tex]:
[tex]$
\text{moles of } \text{O}_2 = \frac{8.0 \, \text{grams of } \text{O}_2}{32.00 \, \text{g/mol}} = 0.25 \, \text{moles of } \text{O}_2
$[/tex]
### Step 4: Use the stoichiometry of the reaction to find moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex]
According to the balanced reaction, 3 moles of [tex]\(\text{O}_2\)[/tex] produce 2 moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex]. Thus, the moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produced can be calculated as:
[tex]$
\text{moles of } \text{Al}_2\text{O}_3 = \left(\frac{2}{3}\right) \times 0.25 \, \text{moles of } \text{O}_2 = 0.16666666666666666 \, \text{moles of } \text{Al}_2\text{O}_3
$[/tex]
### Step 5: Convert the moles of [tex]\(\text{Al}_2\text{O}_3\)[/tex] to mass
Finally, we need to find the mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produced:
[tex]$
\text{mass of } \text{Al}_2\text{O}_3 = \text{moles of } \text{Al}_2\text{O}_3 \times \text{molar mass of } \text{Al}_2\text{O}_3
$[/tex]
[tex]$
\text{mass of } \text{Al}_2\text{O}_3 = 0.16666666666666666 \, \text{moles} \times 101.96 \, \text{g/mol} = 16.993333333333332 \, \text{grams}
$[/tex]
### Conclusion
The mass of aluminum oxide ([tex]\(\text{Al}_2\text{O}_3\)[/tex]) that could form from 8.0 grams of oxygen ([tex]\(\text{O}_2\)[/tex]) with excess aluminum is approximately [tex]\(17.0 \, \text{grams}\)[/tex].
