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Sagot :
To determine the maximum amount of Fe produced during the experiment, we need to consider the balanced chemical equation:
[tex]\[ \text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \][/tex]
We are given the starting amounts of the reactants:
- 3 moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex]
- 5 moles of [tex]\(\text{Al}\)[/tex]
Next, we need to use these amounts to find out how much Fe can be produced.
### 1. Calculate Fe produced from [tex]\(\text{Fe}_2\text{O}_3\)[/tex]
From the balanced equation, we see that 1 mole of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] produces 2 moles of Fe.
Given:
- 3 moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex]
Using the stoichiometric ratio:
[tex]\[ 3 \text{ moles of } \text{Fe}_2\text{O}_3 \times \frac{2 \text{ moles Fe}}{1 \text{ mole } \text{Fe}_2\text{O}_3} = 6 \text{ moles of Fe} \][/tex]
### 2. Calculate Fe produced from Al
From the balanced equation, we see that 2 moles of Al produce 2 moles of Fe.
Given:
- 5 moles of Al
Using the stoichiometric ratio:
[tex]\[ 5 \text{ moles of } \text{Al} \times \frac{2 \text{ moles Fe}}{2 \text{ moles } \text{Al}} = 5 \text{ moles of Fe} \][/tex]
### 3. Determine the limiting reactant and the maximum Fe produced
The production of Fe is limited by the reactant that produces the least amount of Fe. Therefore, the limiting reactant determines the maximum amount of Fe that can be produced.
From the calculations:
- Fe produced from [tex]\(\text{Fe}_2\text{O}_3\)[/tex]: 6 moles
- Fe produced from Al: 5 moles
The limiting reactant here is Al, as it produces the lesser amount of Fe.
Hence, the maximum amount of Fe produced during the experiment is 5 moles.
This conclusion considers the balanced chemical equation and the stoichiometric relationships between reactants and products, ensuring that both reactants are taken into account to find the limiting factor. This clear determination allows us to confirm that 5 moles of Fe is the maximum amount produced given the initial amounts of reactants.
[tex]\[ \text{Fe}_2\text{O}_3 + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al}_2\text{O}_3 \][/tex]
We are given the starting amounts of the reactants:
- 3 moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex]
- 5 moles of [tex]\(\text{Al}\)[/tex]
Next, we need to use these amounts to find out how much Fe can be produced.
### 1. Calculate Fe produced from [tex]\(\text{Fe}_2\text{O}_3\)[/tex]
From the balanced equation, we see that 1 mole of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] produces 2 moles of Fe.
Given:
- 3 moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex]
Using the stoichiometric ratio:
[tex]\[ 3 \text{ moles of } \text{Fe}_2\text{O}_3 \times \frac{2 \text{ moles Fe}}{1 \text{ mole } \text{Fe}_2\text{O}_3} = 6 \text{ moles of Fe} \][/tex]
### 2. Calculate Fe produced from Al
From the balanced equation, we see that 2 moles of Al produce 2 moles of Fe.
Given:
- 5 moles of Al
Using the stoichiometric ratio:
[tex]\[ 5 \text{ moles of } \text{Al} \times \frac{2 \text{ moles Fe}}{2 \text{ moles } \text{Al}} = 5 \text{ moles of Fe} \][/tex]
### 3. Determine the limiting reactant and the maximum Fe produced
The production of Fe is limited by the reactant that produces the least amount of Fe. Therefore, the limiting reactant determines the maximum amount of Fe that can be produced.
From the calculations:
- Fe produced from [tex]\(\text{Fe}_2\text{O}_3\)[/tex]: 6 moles
- Fe produced from Al: 5 moles
The limiting reactant here is Al, as it produces the lesser amount of Fe.
Hence, the maximum amount of Fe produced during the experiment is 5 moles.
This conclusion considers the balanced chemical equation and the stoichiometric relationships between reactants and products, ensuring that both reactants are taken into account to find the limiting factor. This clear determination allows us to confirm that 5 moles of Fe is the maximum amount produced given the initial amounts of reactants.
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