Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.