When aluminum (Al) is added to a silver chloride (AgCl) solution, a single-displacement reaction takes place. In this type of reaction, a more reactive metal displaces a less reactive metal from its compound. The activity series provided helps determine which metals can displace others.
Let's go through the steps to find the products of the reaction:
1. Identify the Reactants:
- Aluminum (Al)
- Silver chloride (AgCl)
2. Refer to the Activity Series:
- The activity series lists metals in order of reactivity from highest to lowest.
- Aluminum (Al) is higher on the activity series than silver (Ag), which means aluminum is more reactive.
3. Write the General Reaction:
- In a single-displacement reaction, the more reactive aluminum will displace the silver in silver chloride.
- The general reaction form: [tex]\( Al + AgCl \rightarrow AlCl_3 + Ag \)[/tex]
4. Determine the Balanced Chemical Equation:
- To displace one silver atom from its chloride compound, aluminum replaces it to form aluminum chloride.
- 3 silver chloride molecules react with 2 aluminum atoms to form 2 aluminum chloride (AlCl3) and 3 silver atoms (Ag).
Here's the balanced chemical equation:
[tex]\[ 2Al + 3AgCl \rightarrow 2AlCl_3 + 3Ag \][/tex]
5. Conclusion:
- The products formed are aluminum chloride (AlCl3) and silver (Ag).
So, when aluminum is added to a silver chloride solution, the products formed are:
[tex]\[ AlCl_3 \text{ and } Ag \][/tex]
Therefore, the answer is [tex]\( \boxed{AlCl_3 \text{ and } Ag} \)[/tex].
