To determine the products of the given reaction, we need to carefully consider the reactants and the potential products formed:
Given reaction:
[tex]\[ \text{CaI}_2 (aq) + \text{Li}_2 \text{CO}_3 (aq) \rightarrow \text{ ??? } \][/tex]
We can hypothesize the possible products by considering a double displacement reaction. In a double displacement reaction, the cations and anions switch partners. Let's analyze the possible product arrangements:
- [tex]\(\text{Ca}^{2+}\)[/tex] from [tex]\(\text{CaI}_2\)[/tex] can pair with [tex]\(\text{CO}_3^{2-}\)[/tex] from [tex]\(\text{Li}_2 \text{CO}_3\)[/tex], forming [tex]\(\text{CaCO}_3\)[/tex].
- [tex]\(\text{Li}^+\)[/tex] from [tex]\(\text{Li}_2 \text{CO}_3\)[/tex] can pair with [tex]\(\text{I}^-\)[/tex] from [tex]\(\text{CaI}_2\)[/tex], forming [tex]\(\text{LiI}\)[/tex].
Thus, the predicted products are:
[tex]\[ \text{LiI}(aq) + \text{CaCO}_3(s) \][/tex]
Now, let's look at the given options:
1. No reaction
2. [tex]\(\text{LiI}(aq) + \text{CaCO}_3(s)\)[/tex]
3. [tex]\(\text{Li}_2 \text{I}_2(aq) + \text{CaCO}_3(aq)\)[/tex]
4. [tex]\(\text{Li}_2 \text{Ca}(aq) + \text{I}_2 \text{CO}_3(s)\)[/tex]
From our analysis, we determined that the correct products are [tex]\(\text{LiI}\)[/tex] in aqueous form and [tex]\(\text{CaCO}_3\)[/tex] as a solid. Hence, the correct option is:
2. [tex]\(\text{LiI}(aq) + \text{CaCO}_3(s)\)[/tex]
