Write the formula unit, total ionic, and net ionic equation for the precipitation reaction in a beaker after Al(ClO4)3, Rb2S, and CsI salts are added.
Please help!
To write the formula unit, total ionic, and net ionic equations for the precipitation reaction between Al(ClO4)3, Rb2S, and CsI salts, we need to identify the ions involved and determine the products formed.
Let's break down each compound and identify their ions:
Al(ClO4)3:
- Al^(3+)
- ClO4^(-)
Rb2S:
- Rb^(+)
- S^(2-)
CsI:
- Cs^(+)
- I^(-)
Now, let's write the formula unit equation by combining the cations and anions to form the products:
Al^(3+) + 3ClO4^(-) + 2Rb^(+) + S^(2-) + 2Cs^(+) + 2I^(-) → Al(ClO4)3 + Rb2S + CsI
Next, we can write the total ionic equation by representing the soluble salts as their individual ions:
Al^(3+) + 3ClO4^(-) + 2Rb^(+) + S^(2-) + 2Cs^(+) + 2I^(-) → Al^(3+) + 3ClO4^(-) + 2Rb^(+) + S^(2-) + 2Cs^(+) + 2I^(-)
In the total ionic equation, we can see that all the ions appear in both the reactant and product side. To write the net ionic equation, we need to eliminate the spectator ions, which are the ions that appear on both sides without undergoing any change. In this case, the spectator ions are Al^(3+), ClO4^(-), Rb^(+), Cs^(+), and I^(-):
S^(2-) → Rb2S + CsI
Thus, the net ionic equation for the precipitation reaction is:
S^(2-) → Rb2S + CsI
To write the formula unit, total ionic, and net ionic equation for the precipitation reaction in a beaker after Al(ClO4)3, Rb2S, and CsI salts are added, we need to determine the products formed when these salts react and the solubility rules for each compound involved.
The first step is to write the balanced equation for the reaction by combining the cation and anion of each compound. Here are the reactants:
Al(ClO4)3 (aqueous) + Rb2S (aqueous) + CsI (aqueous)
Now let's identify the products formed by exchanging the anions and cations:
- In Al(ClO4)3, the cation is Al3+ and the anion is ClO4-. We will pair Al3+ with each of the anions from Rb2S and CsI.
- In Rb2S, the cation is Rb+ and the anion is S2-. We will pair Rb+ with ClO4- from Al(ClO4)3.
- In CsI, the cation is Cs+ and the anion is I-. We will also pair Cs+ with ClO4- from Al(ClO4)3.
Writing the balanced equation:
Al(ClO4)3 (aq) + Rb2S (aq) + CsI (aq) -> Al(ClO4)3 + RbClO4 + CsClO4 + RsI
Now, let's determine the solubility of the compounds:
- Al(ClO4)3: All perchlorates (ClO4-) salts are soluble.
- RbClO4 and CsClO4: All perchlorates (ClO4-) salts are soluble.
- RsI: All sulfides (S2-) salts, except those of group 1A and 2A cations, are insoluble.
Based on the solubility rules, we can see that RbClO4, CsClO4, and Al(ClO4)3 are all soluble and will remain in the solution. However, RsI will undergo a precipitation reaction since sulfides of group 1A cations are insoluble.
Now we can write the formula unit, total ionic, and net ionic equations:
Formula unit equation (molecular equation):
Al(ClO4)3 (aq) + Rb2S (aq) + CsI (aq) -> Al(ClO4)3 + RbClO4 + CsClO4 + RsI (s)
Total ionic equation:
3Al3+ (aq) + 9(ClO4)- (aq) + 2Rb+ (aq) + S2- (aq) + 2Cs+ (aq) + I- (aq) -> 3Al(ClO4)3 (aq) + 2RbClO4 (aq) + 2CsClO4 (aq) + RsI (s)
Net ionic equation:
S2- (aq) + RsI (s) -> RsI (s)
In the net ionic equation, we only include the species that are directly involved in the precipitation reaction, which in this case is the sulfide ion (S2-) and the insoluble compound RsI.
Remember, when writing equations, it is crucial to consider the solubility rules of the compounds to determine if a precipitation reaction occurs or if the compounds remain in solution.