If a precipitation reaction occurs when a solution of one species is added to a solution of the other, enter the formulas and states of all products. Separate by comma. If there is no reaction, enter the formula and state of all ions in solution.
aluminum sulfate + lead(II) nitrate →
To determine the formulas and states of all the products when aluminum sulfate reacts with lead(II) nitrate, we need to identify the ions present in each compound and then use them to form the products.
1. Aluminum sulfate (Al₂(SO₄)₃):
- Aluminum ion (Al³⁺)
- Sulfate ion (SO₄²⁻)
2. Lead(II) nitrate (Pb(NO₃)₂):
- Lead(II) ion (Pb²⁺)
- Nitrate ion (NO₃⁻)
Now, let's figure out the possible products by combining the ions:
The aluminum ion (Al³⁺) has a charge of +3, and the nitrate ion (NO₃⁻) has a charge of -1. To balance the charges, we need three nitrate ions for every aluminum ion. Therefore, the reaction will proceed as follows:
Al₂(SO₄)₃ + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3PbSO₄
The formulas and states of all the products are:
- Aluminum nitrate (Al(NO₃)₃) - aqueous (aq)
- Lead(II) sulfate (PbSO₄) - solid (s)
Therefore, the final answer is:
2Al(NO₃)₃ (aq) + 3PbSO₄ (s)