Answer :
Sure, let's solve this step-by-step.
### 1. Determine the Limiting Reactant:
First, we need to figure out which reactant is the limiting reagent since it will determine the maximum amount of product that can be formed.
Balanced Chemical Equation:
[tex]$ 4 Al (s) + 3 O_2 (g) \rightarrow 2 Al_2O_3 (s) $[/tex]
Given Quantities:
- [tex]\(0.500 \, \text{mol Al}\)[/tex]
- [tex]\(0.500 \, \text{mol O}_2\)[/tex]
1. For Aluminum (Al):
To fully react with [tex]\( 0.500 \)[/tex] moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ 0.500\, \text{mol Al} \times \left(\frac{3 \, \text{mol O}_2}{4 \, \text{mol Al}}\right) = 0.375 \, \text{mol O}_2 \, \text{required} \][/tex]
However, we have [tex]\(0.500 \, \text{mol O}_2\)[/tex] available.
2. For Oxygen (O}_2):
To fully react with [tex]\(0.500 \, \text{mol O}_2\)[/tex]:
[tex]\[ 0.500\, \text{mol O}_2 \times \left(\frac{4 \, \text{mol Al}}{3 \, \text{mol O}_2}\right) = 0.667 \, \text{mol Al} \, \text{required} \][/tex]
However, we have [tex]\(0.500 \, \text{mol Al}\)[/tex] available.
Since [tex]\(0.500 \, \text{mol O}_2\)[/tex] is available and only [tex]\(0.375 \, \text{mol O}_2\)[/tex] is required for [tex]\(0.500 \, \text{mol Al}\)[/tex]:
- [tex]\( \text{Al}\)[/tex] is in excess.
- [tex]\( \text{O}_2\)[/tex] is the limiting reactant.
### 2. Calculate Theoretical Yield of Aluminum Oxide ([tex]\( Al_2O_3 \)[/tex]):
Using the limiting reactant ([tex]\( O_2 \)[/tex]) to calculate the yield of [tex]\( Al_2O_3 \)[/tex]:
[tex]\[ \text{Moles of } Al_2O_3 = 0.500 \, \text{mol O}_2 \times \left(\frac{2 \, \text{mol Al}_2O_3}{3 \, \text{mol O}_2}\right) = 0.333 \, \text{mol Al}_2O_3 \][/tex]
### 3. Convert Moles of [tex]\( Al_2O_3 \)[/tex] to Mass:
Given the molar mass of [tex]\( Al_2O_3 \)[/tex] is [tex]\(101.96 \, \text{g/mol}\)[/tex]:
[tex]\[ \text{Mass of } Al_2O_3 = 0.333 \, \text{mol Al}_2O_3 \times 101.96 \, \text{g/mol} = 33.99 \, \text{g} \][/tex]
### Conclusion:
The theoretical yield of aluminum oxide ([tex]\( Al_2O_3 \)[/tex]) when [tex]\( 0.500 \, \text{mol Al}\)[/tex] reacts with [tex]\( 0.500 \, \text{mol } O_2\)[/tex] is:
- Moles of [tex]\( Al_2O_3\)[/tex]: [tex]\(0.333 \, \text{mol} \)[/tex]
- Mass of [tex]\( Al_2O_3\)[/tex]: [tex]\(33.99 \, \text{g} \)[/tex]
So the maximum amount of aluminum oxide that can be produced under these conditions is [tex]\(33.99 \, \text{grams}\)[/tex].
### 1. Determine the Limiting Reactant:
First, we need to figure out which reactant is the limiting reagent since it will determine the maximum amount of product that can be formed.
Balanced Chemical Equation:
[tex]$ 4 Al (s) + 3 O_2 (g) \rightarrow 2 Al_2O_3 (s) $[/tex]
Given Quantities:
- [tex]\(0.500 \, \text{mol Al}\)[/tex]
- [tex]\(0.500 \, \text{mol O}_2\)[/tex]
1. For Aluminum (Al):
To fully react with [tex]\( 0.500 \)[/tex] moles of [tex]\( \text{Al} \)[/tex]:
[tex]\[ 0.500\, \text{mol Al} \times \left(\frac{3 \, \text{mol O}_2}{4 \, \text{mol Al}}\right) = 0.375 \, \text{mol O}_2 \, \text{required} \][/tex]
However, we have [tex]\(0.500 \, \text{mol O}_2\)[/tex] available.
2. For Oxygen (O}_2):
To fully react with [tex]\(0.500 \, \text{mol O}_2\)[/tex]:
[tex]\[ 0.500\, \text{mol O}_2 \times \left(\frac{4 \, \text{mol Al}}{3 \, \text{mol O}_2}\right) = 0.667 \, \text{mol Al} \, \text{required} \][/tex]
However, we have [tex]\(0.500 \, \text{mol Al}\)[/tex] available.
Since [tex]\(0.500 \, \text{mol O}_2\)[/tex] is available and only [tex]\(0.375 \, \text{mol O}_2\)[/tex] is required for [tex]\(0.500 \, \text{mol Al}\)[/tex]:
- [tex]\( \text{Al}\)[/tex] is in excess.
- [tex]\( \text{O}_2\)[/tex] is the limiting reactant.
### 2. Calculate Theoretical Yield of Aluminum Oxide ([tex]\( Al_2O_3 \)[/tex]):
Using the limiting reactant ([tex]\( O_2 \)[/tex]) to calculate the yield of [tex]\( Al_2O_3 \)[/tex]:
[tex]\[ \text{Moles of } Al_2O_3 = 0.500 \, \text{mol O}_2 \times \left(\frac{2 \, \text{mol Al}_2O_3}{3 \, \text{mol O}_2}\right) = 0.333 \, \text{mol Al}_2O_3 \][/tex]
### 3. Convert Moles of [tex]\( Al_2O_3 \)[/tex] to Mass:
Given the molar mass of [tex]\( Al_2O_3 \)[/tex] is [tex]\(101.96 \, \text{g/mol}\)[/tex]:
[tex]\[ \text{Mass of } Al_2O_3 = 0.333 \, \text{mol Al}_2O_3 \times 101.96 \, \text{g/mol} = 33.99 \, \text{g} \][/tex]
### Conclusion:
The theoretical yield of aluminum oxide ([tex]\( Al_2O_3 \)[/tex]) when [tex]\( 0.500 \, \text{mol Al}\)[/tex] reacts with [tex]\( 0.500 \, \text{mol } O_2\)[/tex] is:
- Moles of [tex]\( Al_2O_3\)[/tex]: [tex]\(0.333 \, \text{mol} \)[/tex]
- Mass of [tex]\( Al_2O_3\)[/tex]: [tex]\(33.99 \, \text{g} \)[/tex]
So the maximum amount of aluminum oxide that can be produced under these conditions is [tex]\(33.99 \, \text{grams}\)[/tex].
