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

Explanation

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

### Step 1: Calculate the mass of AgNO₃ in the solution
Given that the solution is 16.9% AgNO₃, this means there are 16.9 grams of AgNO₃ in 100 mL of solution.

For 50 mL of 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 moles of AgNO₃
The molar mass of AgNO₃ is approximately 170 g/mol.

Using the mass calculated:
\[ \text{Moles of AgNO₃} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{8.45 \text{ g}}{170 \text{ g/mol}} \approx 0.0497 \text{ moles} \]

### Step 3: Calculate the mass of NaCl in the solution
Given that the solution is 5.8% NaCl, this means there are 5.8 grams of NaCl in 100 mL of solution.

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

### Step 4: Calculate the moles of NaCl
The molar mass of NaCl is approximately 58.5 g/mol.

Using the mass calculated:
\[ \text{Moles of NaCl} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.9 \text{ g}}{58.5 \text{ g/mol}} \approx 0.0495 \text{ moles} \]

### Step 5: Determine the limiting reagent
The balanced reaction is:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} \downarrow + \text{NaNO}_3 \]

From the stoichiometry of the reaction, 1 mole of AgNO₃ reacts with 1 mole of NaCl to produce 1 mole of AgCl.

Since the moles of NaCl (0.0495) are slightly less than the moles of AgNO₃ (0.0497), NaCl is the limiting reagent.

### Step 6: Calculate the moles of AgCl formed
Since NaCl is the limiting reagent, the moles of AgCl formed will be equal to the moles of NaCl:
\[ \text{Moles of AgCl} = 0.0495 \text{ moles} \]

### Step 7: Calculate the mass of AgCl
The molar mass of AgCl is approximately 143.3 g/mol.

Using the moles calculated:
\[ \text{Mass of AgCl} = \text{Moles} \times \text{Molar Mass} = 0.0495 \text{ moles} \times 143.3 \text{ g/mol} \approx 7.1 \text{ g} \]

### Final Answer
The mass of the precipitate (AgCl) formed is approximately **7.1 grams**.

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