What is the mass of the precipitate formed when 50 mL of 16.9% solution of `AgNO_(3)` is mixed with 50 mL of 5.8% NaCl solution?

Options

A. (a)`7 g`
B. (b)`14 g`
C. (c )`28 g`
D. (d)`3.5 g`

Explanation

To find the mass of the precipitate formed when 50 mL of 16.9% AgNO₃ solution is mixed with 50 mL of 5.8% NaCl solution, we can follow these steps:

### Step 1: Calculate the mass of AgNO₃ in the solution
The percentage concentration of AgNO₃ is given as 16.9% (weight/volume). This means that in 100 mL of solution, there are 16.9 grams of AgNO₃.

For 50 mL of this solution:
\[ \text{Mass of AgNO₃} = \frac{16.9 \text{ g}}{100 \text{ mL}} \times 50 \text{ mL} = 8.45 \text{ g} \]

### Step 2: Calculate the mass of NaCl in the solution
The percentage concentration of NaCl is given as 5.8% (weight/volume). This means that in 100 mL of solution, there are 5.8 grams of NaCl.

For 50 mL of this solution:
\[ \text{Mass of NaCl} = \frac{5.8 \text{ g}}{100 \text{ mL}} \times 50 \text{ mL} = 2.9 \text{ g} \]

### Step 3: Determine the moles of AgNO₃ and NaCl
Next, we need to calculate the number of moles of AgNO₃ and NaCl.

1. **Molar mass of AgNO₃**:
- Ag: 107.8 g/mol
- N: 14.0 g/mol
- O: 16.0 g/mol × 3 = 48.0 g/mol
- Total = 107.8 + 14.0 + 48.0 = 169.8 g/mol

Now, calculate the moles of AgNO₃:
\[ \text{Moles of AgNO₃} = \frac{8.45 \text{ g}}{169.8 \text{ g/mol}} \approx 0.0498 \text{ mol} \]

2. **Molar mass of NaCl**:
- Na: 23.0 g/mol
- Cl: 35.5 g/mol
- Total = 23.0 + 35.5 = 58.5 g/mol

Now, calculate the moles of NaCl:
\[ \text{Moles of NaCl} = \frac{2.9 \text{ g}}{58.5 \text{ g/mol}} \approx 0.0497 \text{ mol} \]

### Step 4: Identify the limiting reagent
The balanced chemical reaction is:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \]
From the stoichiometry, 1 mole of AgNO₃ reacts with 1 mole of NaCl.

Since both moles are approximately equal (0.0498 mol of AgNO₃ and 0.0497 mol of NaCl), we can conclude that both reactants will be consumed completely.

### Step 5: Calculate the mass of AgCl formed
Now, we need to calculate the mass of AgCl formed. The molar mass of AgCl is:
- Ag: 107.8 g/mol
- Cl: 35.5 g/mol
- Total = 107.8 + 35.5 = 143.3 g/mol

Using the moles of either reactant (since they are in a 1:1 ratio), we can find the mass of AgCl produced:
\[ \text{Mass of AgCl} = \text{Moles of AgCl} \times \text{Molar mass of AgCl} = 0.0497 \text{ mol} \times 143.3 \text{ g/mol} \approx 7.13 \text{ g} \]

### Final Answer
The mass of the precipitate (AgCl) formed is approximately **7 grams**.

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