Let's break down the given chemical reaction and analyze the quantities involved to determine the accurate observation when the reaction stops.
The reaction given is:
[tex]\[2 A_l + 3 CuCl_2 \rightarrow 3 Cu + 2 AlCl_3\][/tex]
The molar masses of the substances involved are approximately:
- Aluminum, [tex]\(Al\)[/tex]: 27 g/mol
- Copper (II) chloride, [tex]\(CuCl_2\)[/tex]: 134.5 g/mol
- Copper, [tex]\(Cu\)[/tex]: 63.5 g/mol
Our goal is to determine how much copper will be formed and what will be left over (either aluminum or copper chloride):
1. Determine the moles of aluminum and copper (II) chloride:
Given quantities:
- Aluminum: 0.25 grams
- Copper (II) chloride: 0.40 grams
Moles of aluminum:
[tex]\[
\text{Moles of } Al = \frac{0.25 \text{ grams}}{27 \text{ g/mol}} \approx 0.00926 \text{ moles}
\][/tex]
Moles of copper (II) chloride:
[tex]\[
\text{Moles of } CuCl_2 = \frac{0.40 \text{ grams}}{134.5 \text{ g/mol}} \approx 0.00297 \text{ moles}
\][/tex]
2. Determine the limiting reagent:
The balanced equation shows the stoichiometric ratios:
[tex]\[
2 \text{ moles } Al : 3 \text{ moles } CuCl_2
\][/tex]
For 0.00926 moles of [tex]\(Al\)[/tex], the required moles of [tex]\(CuCl_2\)[/tex] are:
[tex]\[
\text{Required moles of } CuCl_2 = \frac{3}{2} \times 0.00926 \approx 0.01389 \text{ moles}
\][/tex]
Since we only have 0.00297 moles of [tex]\(CuCl_2\)[/tex], [tex]\(CuCl_2\)[/tex] is the limiting reagent.
3. Calculate the amount of copper formed:
From the stoichiometry of the reaction:
[tex]\[
3 \text{ moles of } CuCl_2 \rightarrow 3 \text{ moles of } Cu
\][/tex]
Therefore, 0.00297 moles of [tex]\(CuCl_2\)[/tex] will produce an equal amount of moles of copper:
[tex]\[
\text{Moles of } Cu = 0.00297 \text{ moles}
\][/tex]
Converting moles of [tex]\(Cu\)[/tex] to grams:
[tex]\[
\text{Grams of } Cu = 0.00297 \times 63.5 \approx 0.18895 \text{ grams} \approx 0.19 \text{ grams}
\][/tex]
So, about 0.20 grams of copper will be formed (closest approximation).
4. Determine what is left when the reaction stops:
- Since copper (II) chloride [tex]\(CuCl_2\)[/tex] was the limiting reagent, it will be fully consumed.
- Some aluminum will be left unreacted since there was more than needed for the reaction with 0.00297 moles of [tex]\(CuCl_2\)[/tex].
Therefore, the correct observation when the reaction stops is:
About 0.20 grams of copper is formed, and some aluminum is left in the reaction mixture.
