To solve this problem, we need to determine how many grams of oxygen gas ([tex]\(O_2\)[/tex]) are needed to completely react with 9.3 moles of aluminum ([tex]\(Al\)[/tex]) based on the given chemical reaction:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
Here's a step-by-step solution to find the answer:
### Step 1: Write Down Given Information
- We are given 9.3 moles of aluminum ([tex]\(9.3\)[/tex] mol Al).
- The balanced chemical equation is:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
### Step 2: Determine the Mole Ratio
From the balanced equation, 4 moles of [tex]\( \text{Al} \)[/tex] react with 3 moles of [tex]\( \text{O}_2 \)[/tex]. Thus, the mole ratio of [tex]\( \text{Al} \)[/tex] to [tex]\( \text{O}_2 \)[/tex] is 4:3.
### Step 3: Calculate the Required Moles of [tex]\( \text{O}_2 \)[/tex]
Using the mole ratio, we can find how many moles of [tex]\( \text{O}_2 \)[/tex] are needed for 9.3 moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ \text{Moles of } \text{O}_2 = \left(\frac{3}{4}\right) \times 9.3 \text{ mol Al} = 6.975 \text{ moles of } \text{O}_2 \][/tex]
### Step 4: Find the Molar Mass of [tex]\( \text{O}_2 \)[/tex]
The molar mass of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \text{Molar mass of } \text{O}_2 = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol} \][/tex]
### Step 5: Calculate the Grams of [tex]\( \text{O}_2 \)[/tex] Needed
Now we can find the mass of [tex]\( \text{O}_2 \)[/tex] needed by multiplying the moles of [tex]\( \text{O}_2 \)[/tex] by its molar mass:
[tex]\[ \text{Grams of } \text{O}_2 \text{ needed} = 6.975 \text{ moles} \times 32 \text{ g/mol} = 223.2 \text{ grams of } \text{O}_2 \][/tex]
### Conclusion
- The unknown in the problem is the grams of [tex]\( O_2 \)[/tex] needed.
- Given: [tex]\( 9.3 \)[/tex] mol [tex]\( Al \)[/tex]
Therefore, the number of grams of oxygen gas ([tex]\( O_2 \)[/tex]) needed to completely react with [tex]\( 9.3 \)[/tex] moles of aluminum is:
[tex]\[ 223.2 \text{ grams } O_2 \][/tex]
