How many moles of AgCI will be formed if 10 g each of KCI and NaCI react with excess of silver nitrate?

Explanation

To solve the problem of how many moles of AgCl will be formed when 10 g each of KCl and NaCl react with excess silver nitrate, we will follow these steps:

### Step 1: Write the chemical equations for the reactions.
1. **For KCl:**
\[ \text{KCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{KNO}_3 \]

2. **For NaCl:**
\[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \]

### Step 2: Calculate the molar masses of KCl and NaCl.
- **Molar mass of KCl:**
- K (Potassium) = 39.1 g/mol
- Cl (Chlorine) = 35.5 g/mol
- Total = 39.1 + 35.5 = 74.6 g/mol

- **Molar mass of NaCl:**
- Na (Sodium) = 23.0 g/mol
- Cl (Chlorine) = 35.5 g/mol
- Total = 23.0 + 35.5 = 58.5 g/mol

### Step 3: Determine the number of moles of KCl and NaCl.
- **Moles of KCl:**
\[ \text{Moles of KCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{74.6 \text{ g/mol}} \approx 0.134 \text{ moles} \]

- **Moles of NaCl:**
\[ \text{Moles of NaCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{58.5 \text{ g/mol}} \approx 0.171 \text{ moles} \]

### Step 4: Determine the amount of AgCl produced from each salt.
- **From KCl:**
- According to the reaction, 1 mole of KCl produces 1 mole of AgCl.
- Therefore, 0.134 moles of KCl will produce 0.134 moles of AgCl.

- **From NaCl:**
- Similarly, 1 mole of NaCl produces 1 mole of AgCl.
- Therefore, 0.171 moles of NaCl will produce 0.171 moles of AgCl.

### Step 5: Identify the limiting reagent.
- Since KCl produces fewer moles of AgCl (0.134 moles) compared to NaCl (0.171 moles), KCl is the limiting reagent.

### Step 6: Calculate the total moles of AgCl formed.
- The total moles of AgCl formed will be equal to the moles produced from KCl since it is the limiting reagent:
\[ \text{Total moles of AgCl} = 0.134 \text{ moles} \]

### Final Answer:
The total number of moles of AgCl formed is approximately **0.134 moles**.

---