To solve this problem, we'll use the GRESA format (Given, Required, Equation, Solution, Answer). Let's break it down step by step.

1. Given:

• Mass of AgNO₃ = 25.0 g
• Mass of FeCl₃ = 45.0 g
• Molar masses:
  • AgNO₃ = 169.87 g/mol
  • FeCl₃ = 162.2 g/mol
  • AgCl = 143.32 g/mol

2. Required:

We need to find:

• a. The limiting reagent and excess reagent.
• b. The mass of AgCl formed.
• c. The amount of excess reactant remaining after the reaction.

3. Equation:

The balanced chemical equation is:

$3 \text{AgNO₃} + \text{FeCl₃} \rightarrow 3 \text{AgCl} + \text{Fe(NO₃)₃}$

This tells us that 3 moles of AgNO₃ react with 1 mole of FeCl₃ to produce 3 moles of AgCl.

4. Solution:

Step 1: Calculate the moles of each reactant.

• Moles of AgNO₃: $\text{Moles of AgNO₃} = \frac{25.0 \, \text{g}}{169.87 \, \text{g/mol}} \approx 0.1472 \, \text{mol}$

• Moles of FeCl₃: $\text{Moles of FeCl₃} = \frac{45.0 \, \text{g}}{162.2 \, \text{g/mol}} \approx 0.2774 \, \text{mol}$

Step 2: Determine the limiting reactant.

From the balanced equation, 3 moles of AgNO₃ react with 1 mole of FeCl₃, so we compare their molar ratio.

For AgNO₃:
We need $\frac{1}{3} \times 0.1472 = 0.0491$ moles of FeCl₃ to react with 0.1472 moles of AgNO₃.

For FeCl₃:
We need $3 \times 0.2774 = 0.832$ moles of AgNO₃ to react with 0.2774 moles of FeCl₃.

Since we have only 0.1472 moles of AgNO₃ (and need 0.832), AgNO₃ is the limiting reagent, and FeCl₃ is in excess.

Step 3: Calculate the amount of AgCl formed.

From the balanced equation, 3 moles of AgNO₃ produce 3 moles of AgCl. Therefore, the moles of AgCl formed are the same as the moles of AgNO₃.

$\text{Moles of AgCl} = \text{Moles of AgNO₃} = 0.1472 \, \text{mol}$

Now, calculate the mass of AgCl:

$\text{Mass of AgCl} = 0.1472 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 21.10 \, \text{g}$

Step 4: Calculate the excess FeCl₃ remaining.

We already know that 0.0491 moles of FeCl₃ are required to react with the 0.1472 moles of AgNO₃.

Moles of FeCl₃ remaining: $\text{Moles of FeCl₃ remaining} = 0.2774 \, \text{mol} - 0.0491 \, \text{mol} = 0.2283 \, \text{mol}$

Convert the remaining moles of FeCl₃ to grams: $\text{Mass of FeCl₃ remaining} = 0.2283 \, \text{mol} \times 162.2 \, \text{g/mol} \approx 37.0 \, \text{g}$

5. Answer:

• a. Limiting reagent: AgNO₃, Excess reagent: FeCl₃
• b. Mass of AgCl formed: 21.10 g
• c. Mass of FeCl₃ remaining: 37.0 g

Would you like any more details or have further questions? Here are five related questions:

1. What happens if the amount of AgNO₃ is increased?
2. How do you determine the molar mass of compounds?
3. Why does the limiting reagent determine the amount of product formed?
4. How would the reaction change if FeCl₃ were the limiting reagent?
5. What is the significance of a balanced chemical equation?

Tip: Always check the stoichiometric ratio from the balanced equation to find the limiting reagent.