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Sagot :
Sure, let's determine the theoretical yield of copper (Cu) from the given balanced chemical equation:
[tex]\[ 3 \text{CuCl}_2 + 2 \text{Al} \rightarrow 2 \text{AlCl}_3 + 3 \text{Cu} \][/tex]
Here, we are given that aluminum ([tex]\(\text{Al}\)[/tex]) is the limiting reactant.
1. Identify the molar ratio:
From the balanced equation, we can see the molar ratio between aluminum and copper. For every 2 moles of aluminum, 3 moles of copper are produced.
2. Express the molar ratio:
The molar ratio between aluminum and copper is:
[tex]\[ 2 \text{ mol Al} : 3 \text{ mol Cu} \][/tex]
3. Determine the theoretical yield:
If 2 moles of aluminum produce 3 moles of copper, then we need to find out how many moles of copper are produced by 1 mole of aluminum (since aluminum is the limiting reactant here).
To find this, we can set up the following proportion:
[tex]\[ \frac{3 \text{ mol Cu}}{2 \text{ mol Al}} \][/tex]
Therefore, for 1 mole of aluminum, the amount of copper produced would be:
[tex]\[ \left(\frac{3 \text{ mol Cu}}{2 \text{ mol Al}}\right) \times 1 \text{ mol Al} = 1.5 \text{ mol Cu} \][/tex]
Thus, the theoretical yield of copper is:
[tex]\[ \boxed{1.5} \text{ moles} \][/tex]
So, the theoretical yield of copper when aluminum is the limiting reactant is [tex]\(1.5\)[/tex] moles.
[tex]\[ 3 \text{CuCl}_2 + 2 \text{Al} \rightarrow 2 \text{AlCl}_3 + 3 \text{Cu} \][/tex]
Here, we are given that aluminum ([tex]\(\text{Al}\)[/tex]) is the limiting reactant.
1. Identify the molar ratio:
From the balanced equation, we can see the molar ratio between aluminum and copper. For every 2 moles of aluminum, 3 moles of copper are produced.
2. Express the molar ratio:
The molar ratio between aluminum and copper is:
[tex]\[ 2 \text{ mol Al} : 3 \text{ mol Cu} \][/tex]
3. Determine the theoretical yield:
If 2 moles of aluminum produce 3 moles of copper, then we need to find out how many moles of copper are produced by 1 mole of aluminum (since aluminum is the limiting reactant here).
To find this, we can set up the following proportion:
[tex]\[ \frac{3 \text{ mol Cu}}{2 \text{ mol Al}} \][/tex]
Therefore, for 1 mole of aluminum, the amount of copper produced would be:
[tex]\[ \left(\frac{3 \text{ mol Cu}}{2 \text{ mol Al}}\right) \times 1 \text{ mol Al} = 1.5 \text{ mol Cu} \][/tex]
Thus, the theoretical yield of copper is:
[tex]\[ \boxed{1.5} \text{ moles} \][/tex]
So, the theoretical yield of copper when aluminum is the limiting reactant is [tex]\(1.5\)[/tex] moles.
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