17 gm `AgNO_3` is treated with 25 gm HCl. What is the mass of AgCl formed ?

Options

A. 18
B. 98.3
C. 24.7
D. 14.35

Answer

To solve the problem of how much AgCl is formed when 17 g of AgNO3 is treated with 25 g of HCl, we will follow these steps:

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

### Step 2: Calculate the molar masses
- Molar mass of AgNO3:
- Ag = 107.87 g/mol
- N = 14.01 g/mol
- O3 = 3 × 16.00 g/mol = 48.00 g/mol
- Total = 107.87 + 14.01 + 48.00 = 169.88 g/mol (approximately 170 g/mol)

- Molar mass of HCl:
- H = 1.01 g/mol
- Cl = 35.45 g/mol
- Total = 1.01 + 35.45 = 36.46 g/mol (approximately 36.5 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: Determine the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of AgNO3 reacts with 1 mole of HCl to produce 1 mole of AgCl.

### Step 4: Calculate the moles of each reactant
- Moles of AgNO3:
\[ \text{Moles of AgNO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{17 \text{ g}}{170 \text{ g/mol}} = 0.1 \text{ moles} \]

- Moles of HCl:
\[ \text{Moles of HCl} = \frac{25 \text{ g}}{36.5 \text{ g/mol}} \approx 0.685 \text{ moles} \]

### Step 5: Identify the limiting reagent
Since the stoichiometry is 1:1, we compare the moles:
- 0.1 moles of AgNO3 requires 0.1 moles of HCl.
- We have 0.685 moles of HCl available.

Thus, AgNO3 is the limiting reagent because it will be completely consumed before HCl.

### Step 6: Calculate the mass of AgCl produced
Using the moles of the limiting reagent (AgNO3):
- From the balanced equation, 1 mole of AgNO3 produces 1 mole of AgCl.
- Therefore, 0.1 moles of AgNO3 will produce 0.1 moles of AgCl.

Now, calculate the mass of AgCl:
\[ \text{Mass of AgCl} = \text{moles} \times \text{molar mass} = 0.1 \text{ moles} \times 143.5 \text{ g/mol} = 14.35 \text{ g} \]

### Final Answer
The mass of AgCl formed is **14.35 grams**.

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