The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4 g of `AgNO_3` is [Atomic mass of Ag = 108, Atomic mass of Na = 23]

Options

A. 5.74 g
B. 1.17 g
C. 2.87 g
D. 6.8 g

Explanation

To determine the mass of AgCl precipitated when 11.70 g of NaCl is added to a solution containing 3.4 g of AgNO3, we will follow these steps:

### Step 1: Write the Balanced Chemical Equation
The reaction between NaCl and AgNO3 can be represented as:
\[ \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \]

### Step 2: Calculate Moles of NaCl
To find the moles of NaCl, we need its molar mass:
- Molar mass of Na = 23 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Now, calculate the moles of NaCl:
\[ \text{Moles of NaCl} = \frac{\text{mass of NaCl}}{\text{molar mass of NaCl}} = \frac{11.70 \text{ g}}{58.5 \text{ g/mol}} \approx 0.200 \text{ mol} \]

### Step 3: Calculate Moles of AgNO3
Next, we calculate the moles of AgNO3. The molar mass of AgNO3 is:
- Molar mass of Ag = 108 g/mol
- Molar mass of N = 14 g/mol
- Molar mass of O = 16 g/mol (3 O atoms)
- Molar mass of AgNO3 = 108 + 14 + (3 \times 16) = 170 g/mol

Now, calculate the moles of AgNO3:
\[ \text{Moles of AgNO}_3 = \frac{\text{mass of AgNO}_3}{\text{molar mass of AgNO}_3} = \frac{3.4 \text{ g}}{170 \text{ g/mol}} \approx 0.020 \text{ mol} \]

### Step 4: Identify the Limiting Reagent
From the balanced equation, we see that 1 mole of NaCl reacts with 1 mole of AgNO3 to produce 1 mole of AgCl. We have:
- Moles of NaCl = 0.200 mol
- Moles of AgNO3 = 0.020 mol

Since AgNO3 has fewer moles, it is the limiting reagent.

### Step 5: Calculate Moles of AgCl Produced
According to the stoichiometry of the reaction:
\[ \text{Moles of AgCl produced} = \text{Moles of AgNO}_3 = 0.020 \text{ mol} \]

### Step 6: Calculate the Mass of AgCl
Now, we need to calculate the mass of AgCl produced. The molar mass of AgCl is:
- Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Now, calculate the mass of AgCl:
\[ \text{Mass of AgCl} = \text{moles of AgCl} \times \text{molar mass of AgCl} = 0.020 \text{ mol} \times 143.5 \text{ g/mol} \approx 2.87 \text{ g} \]

### Final Answer
The mass of AgCl precipitated is approximately **2.87 g**.

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