What weight of `AgCl` will be precipitated when a solution containing `4.77 g NaCl` is added to a solution of `5.77 g` of `AgNO_(3)`.

Explanation

To solve the problem of how much AgCl will be precipitated when 4.77 g of NaCl is added to 5.77 g of AgNO3, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) can be represented as:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \]

### Step 2: Calculate the molar masses
- **Molar mass of AgNO3**:
- Ag = 107.87 g/mol
- N = 14.01 g/mol
- O = 16.00 g/mol × 3 = 48.00 g/mol
- Total = 107.87 + 14.01 + 48.00 = 169.88 g/mol (approximately 170 g/mol)

- **Molar mass of NaCl**:
- Na = 22.99 g/mol
- Cl = 35.45 g/mol
- Total = 22.99 + 35.45 = 58.44 g/mol (approximately 58.5 g/mol)

- **Molar mass of AgCl**:
- Ag = 107.87 g/mol
- Cl = 35.45 g/mol
- Total = 107.87 + 35.45 = 143.32 g/mol (approximately 143.5 g/mol)

### Step 3: Calculate the number of moles of AgNO3 and NaCl
- **Moles of AgNO3**:
\[ \text{Moles of AgNO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.77 \, \text{g}}{170 \, \text{g/mol}} \approx 0.0339 \, \text{mol} \]

- **Moles of NaCl**:
\[ \text{Moles of NaCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.77 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.0815 \, \text{mol} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.

- We have 0.0339 moles of AgNO3 and 0.0815 moles of NaCl.
- Since AgNO3 is present in a smaller amount, it is the limiting reagent.

### Step 5: Calculate the moles of AgCl produced
Since the reaction produces AgCl in a 1:1 ratio with AgNO3, the moles of AgCl produced will be equal to the moles of AgNO3:
\[ \text{Moles of AgCl} = 0.0339 \, \text{mol} \]

### Step 6: Calculate the mass of AgCl produced
Using the molar mass of AgCl:
\[ \text{Mass of AgCl} = \text{moles} \times \text{molar mass} = 0.0339 \, \text{mol} \times 143.5 \, \text{g/mol} \approx 4.87 \, \text{g} \]

### Conclusion
The weight of AgCl that will be precipitated is approximately **4.87 g**.

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