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Let's break down the steps required to solve this problem systematically:

### Given:
- **Potassium phosphate (K₃PO₄)** mass = 70.5 mg = 0.0705 g
- **Silver nitrate (AgNO₃)** solution = 15.0 mL of 0.050 M solution
- Molar masses: 
  - **N = 14, O = 16, P = 31, K = 39.1, Ag = 107.9** (all g/mol)

We will solve this step by step.

### a. Persamaan keseimbangan molekul (Balanced Molecular Equation)

The reaction involves **potassium phosphate (K₃PO₄)** and **silver nitrate (AgNO₃)**. When these two compounds react, silver phosphate (Ag₃PO₄) forms as a precipitate, and potassium nitrate (KNO₃) remains in the solution.

The balanced molecular equation is:
\[
\text{K₃PO₄ (aq) + 3AgNO₃ (aq) → Ag₃PO₄ (s) + 3KNO₃ (aq)}
\]

### b. Persamaan ion lengkap (Complete Ionic Equation)

In an aqueous solution, potassium phosphate and silver nitrate dissociate into their respective ions. The complete ionic equation is:
\[
\text{3K⁺ (aq) + PO₄³⁻ (aq) + 3Ag⁺ (aq) + 3NO₃⁻ (aq) → Ag₃PO₄ (s) + 3K⁺ (aq) + 3NO₃⁻ (aq)}
\]

### c. Persamaan ion bersih (Net Ionic Equation)

To obtain the net ionic equation, we cancel out the spectator ions (K⁺ and NO₃⁻) that appear on both sides. The net ionic equation is:
\[
\text{PO₄³⁻ (aq) + 3Ag⁺ (aq) → Ag₃PO₄ (s)}
\]

### d. Pereaksi Pembatas (Limiting Reactant)

To determine the limiting reactant, we calculate the moles of each reactant:

1. **Moles of potassium phosphate (K₃PO₄):**
   \[
   \text{Molar mass of K₃PO₄} = 3(39.1) + 31 + 4(16) = 212.27 \, \text{g/mol}
   \]
   \[
   \text{Moles of K₃PO₄} = \frac{0.0705 \, \text{g}}{212.27 \, \text{g/mol}} = 3.32 \times 10^{-4} \, \text{mol}
   \]

2. **Moles of silver nitrate (AgNO₃):**
   \[
   \text{Moles of AgNO₃} = 0.050 \, \text{M} \times 0.015 \, \text{L} = 7.5 \times 10^{-4} \, \text{mol}
   \]

From the balanced equation, 1 mole of K₃PO₄ reacts with 3 moles of AgNO₃. Therefore, the required moles of AgNO₃ for the available K₃PO₄ are:
\[
\text{Required moles of AgNO₃} = 3 \times 3.32 \times 10^{-4} = 9.96 \times 10^{-4} \, \text{mol}
\]
Since we only have 7.5 × 10⁻⁴ mol of AgNO₃, silver nitrate (AgNO₃) is the limiting reactant.

### e. Hitung Hasil Teoritis (Theoretical Yield of Precipitate)

The theoretical yield of the precipitate (Ag₃PO₄) can be calculated from the moles of the limiting reactant, AgNO₃.

1. **Moles of Ag₃PO₄ formed:**
   From the balanced equation, 3 moles of AgNO₃ produce 1 mole of Ag₃PO₄. Therefore:
   \[
   \text{Moles of Ag₃PO₄} = \frac{7.5 \times 10^{-4}}{3} = 2.5 \times 10^{-4} \, \text{mol}
   \]

2. **Molar mass of Ag₃PO₄:**
   \[
   \text{Molar mass of Ag₃PO₄} = 3(107.9) + 31 + 4(16) = 418.58 \, \text{g/mol}
   \]

3. **Mass of Ag₃PO₄ formed:**
   \[
   \text{Mass of Ag₃PO₄} = 2.5 \times 10^{-4} \, \text{mol} \times 418.58 \, \text{g/mol} = 0.1046 \, \text{g}
   \]

The theoretical mass of Ag₃PO₄ formed is **0.1046 g**.

---

### Do you want further details or have any questions?

Here are 5 follow-up questions to expand on this topic:

1. How would the mass of the precipitate change if the concentration of silver nitrate increased?
2. What is the role of spectator ions in ionic reactions, and how do they affect the overall reaction?
3. How would you experimentally determine the actual yield of the precipitate?
4. Can limiting reactants always be identified using stoichiometry alone, or are there other factors to consider?
5. How would the presence of impurities in potassium phosphate affect the theoretical yield?

#### Tip:
Always ensure your solution units (e.g., moles, grams) match before performing stoichiometric calculations to avoid errors.

Sampel 70,5 mg kalium fosfat ditambahkan ke 15,0 mL 0,050 M perak nitrat menghasilkan pembentukan endapan. Tulis untuk reaksi tersebut: (a) persamaan keseimbangan molekul, (b) persamaan ion lengkap, (c) persamaan ion bersih, (d) pereaksi pembatas, (e) hitung hasil teoritis, dalam gram, dari endapan yang terbentuk.

Solve a stoichiometry problem involving potassium phosphate and silver nitrate to determine the limiting reactant, balanced molecular and ionic equations, and the theoretical yield of silver phosphate precipitate. Suitable for high school students learning chemistry.

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