What is the mass of the precipitate formed when 50 mL of 16.9 % (w//v ) solution of `Ag NO_(3)` is mixed with 50 mL of 5.8 % (w/v) NaCl solutions (Ag = 107.3 ,N=14,O=16 ,Na=23 ,Cl=35.5)

Options

A. 7g
B. 14g
C. 28g
D. `3.5g`

Answer

To find the mass of the precipitate formed when 50 mL of 16.9% (w/v) AgNO₃ solution is mixed with 50 mL of 5.8% (w/v) NaCl solution, we will follow these steps:

### Step 1: Calculate the mass of AgNO₃ in the solution
The percentage (w/v) means grams of solute per 100 mL of solution.

For AgNO₃:
- Given concentration = 16.9% (w/v)
- Volume of solution = 50 mL

\[ \text{Mass of AgNO}_3 = \left(\frac{16.9 \text{ g}}{100 \text{ mL}}\right) \times 50 \text{ mL} = 8.45 \text{ g} \]

### Step 2: Calculate the number of moles of AgNO₃
To find the number of moles, use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of AgNO₃ is calculated as follows:
- Ag = 107.3 g/mol
- N = 14 g/mol
- O = 16 g/mol × 3 = 48 g/mol
- Total = 107.3 + 14 + 48 = 169.3 g/mol

Now, calculate the number of moles of AgNO₃:

\[ \text{Number of moles of AgNO}_3 = \frac{8.45 \text{ g}}{169.3 \text{ g/mol}} \approx 0.0499 \text{ moles} \]

### Step 3: Calculate the mass of NaCl in the solution
For NaCl:
- Given concentration = 5.8% (w/v)
- Volume of solution = 50 mL

\[ \text{Mass of NaCl} = \left(\frac{5.8 \text{ g}}{100 \text{ mL}}\right) \times 50 \text{ mL} = 2.9 \text{ g} \]

### Step 4: Calculate the number of moles of NaCl
The molar mass of NaCl is:
- Na = 23 g/mol
- Cl = 35.5 g/mol
- Total = 23 + 35.5 = 58.5 g/mol

Now, calculate the number of moles of NaCl:

\[ \text{Number of moles of NaCl} = \frac{2.9 \text{ g}}{58.5 \text{ g/mol}} \approx 0.0496 \text{ moles} \]

### Step 5: Determine the limiting reagent
The reaction between AgNO₃ and NaCl is:

\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \]

From the stoichiometry of the reaction, 1 mole of AgNO₃ reacts with 1 mole of NaCl to produce 1 mole of AgCl. Since both reactants are present in nearly equal amounts (0.0499 moles of AgNO₃ and 0.0496 moles of NaCl), NaCl is the limiting reagent.

### Step 6: Calculate the mass of AgCl precipitate formed
The number of moles of AgCl formed will be equal to the number of moles of the limiting reagent (NaCl):

\[ \text{Number of moles of AgCl} = 0.0496 \text{ moles} \]

The molar mass of AgCl is:
- Ag = 107.3 g/mol
- Cl = 35.5 g/mol
- Total = 107.3 + 35.5 = 142.8 g/mol

Now, calculate the mass of AgCl:

\[ \text{Mass of AgCl} = \text{Number of moles} \times \text{Molar mass} = 0.0496 \text{ moles} \times 142.8 \text{ g/mol} \approx 7.08 \text{ g} \]

### Final Answer
The mass of the precipitate (AgCl) formed is approximately **7.08 grams**.

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