Answer :
Let's solve the problem step-by-step to determine how many moles of oxygen gas [tex]\((O_2)\)[/tex] are needed to completely react with [tex]\(54.0 \, \text{g}\)[/tex] of aluminum.
### Given Data:
1. Mass of aluminum (Al): [tex]\(54.0 \, \text{g}\)[/tex]
2. Molar mass of aluminum (Al): [tex]\(26.98 \, \text{g/mol}\)[/tex]
3. The balanced chemical equation:
[tex]\[ 4 \, \text{Al} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Al}_2\text{O}_3 \][/tex]
### Step 1: Calculate the moles of aluminum ([tex]\(\text{Al}\)[/tex])
Using the molar mass of aluminum, we can determine the number of moles of aluminum in [tex]\(54.0 \, \text{g}\)[/tex]:
[tex]\[ \text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{54.0 \, \text{g}}{26.98 \, \text{g/mol}} \approx 2.001 \][/tex]
### Step 2: Use the molar ratio from the balanced equation
From the balanced chemical equation, the molar ratio of aluminum to oxygen gas ([tex]\(O_2\)[/tex]) is:
[tex]\[ \frac{4 \, \text{mol Al}}{3 \, \text{mol O}_2} \][/tex]
So for every 4 moles of Al, 3 moles of [tex]\(O_2\)[/tex] are needed. We use this ratio to determine the moles of [tex]\(O_2\)[/tex] needed for the moles of Al calculated:
[tex]\[ \text{Moles of } O_2 = \text{Moles of Al} \times \frac{3 \, \text{mol O}_2}{4 \, \text{mol Al}} = 2.001 \, \text{mol Al} \times \frac{3}{4} \approx 1.501 \][/tex]
### Conclusion
The number of moles of oxygen gas ([tex]\(O_2\)[/tex]) needed to completely react with [tex]\(54.0 \, \text{g}\)[/tex] of aluminum is approximately [tex]\(1.501\)[/tex] moles.
### Given Data:
1. Mass of aluminum (Al): [tex]\(54.0 \, \text{g}\)[/tex]
2. Molar mass of aluminum (Al): [tex]\(26.98 \, \text{g/mol}\)[/tex]
3. The balanced chemical equation:
[tex]\[ 4 \, \text{Al} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Al}_2\text{O}_3 \][/tex]
### Step 1: Calculate the moles of aluminum ([tex]\(\text{Al}\)[/tex])
Using the molar mass of aluminum, we can determine the number of moles of aluminum in [tex]\(54.0 \, \text{g}\)[/tex]:
[tex]\[ \text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{54.0 \, \text{g}}{26.98 \, \text{g/mol}} \approx 2.001 \][/tex]
### Step 2: Use the molar ratio from the balanced equation
From the balanced chemical equation, the molar ratio of aluminum to oxygen gas ([tex]\(O_2\)[/tex]) is:
[tex]\[ \frac{4 \, \text{mol Al}}{3 \, \text{mol O}_2} \][/tex]
So for every 4 moles of Al, 3 moles of [tex]\(O_2\)[/tex] are needed. We use this ratio to determine the moles of [tex]\(O_2\)[/tex] needed for the moles of Al calculated:
[tex]\[ \text{Moles of } O_2 = \text{Moles of Al} \times \frac{3 \, \text{mol O}_2}{4 \, \text{mol Al}} = 2.001 \, \text{mol Al} \times \frac{3}{4} \approx 1.501 \][/tex]
### Conclusion
The number of moles of oxygen gas ([tex]\(O_2\)[/tex]) needed to completely react with [tex]\(54.0 \, \text{g}\)[/tex] of aluminum is approximately [tex]\(1.501\)[/tex] moles.
