Let's break down the problem step by step:

Step 1: Determine the limiting reactant

The balanced chemical equation is: $\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$

We are given:

• Volume of AgNO₃ solution = 74.0 mL (or 0.074 L)
• Molarity of AgNO₃ = 0.200 M
• Volume of NaCl solution = 195.0 mL (or 0.195 L)
• Molarity of NaCl = 0.351 M

1. Calculate the moles of AgNO₃: $\text{Moles of AgNO}_3 = Molarity \times Volume$ $\text{Moles of AgNO}_3 = 0.200 \, \text{M} \times 0.074 \, \text{L} = 0.0148 \, \text{mol}$

2. Calculate the moles of NaCl: $\text{Moles of NaCl} = Molarity \times Volume$ $\text{Moles of NaCl} = 0.351 \, \text{M} \times 0.195 \, \text{L} = 0.068445 \, \text{mol}$

Step 2: Determine the limiting reactant

The reaction is 1:1, meaning one mole of AgNO₃ reacts with one mole of NaCl. Since we have fewer moles of AgNO₃ (0.0148 mol) than NaCl (0.068445 mol), AgNO₃ is the limiting reactant.

Step 3: Calculate the mass of AgCl produced

From the balanced equation, 1 mole of AgNO₃ produces 1 mole of AgCl. Thus, the moles of AgCl produced will be the same as the moles of AgNO₃ used, which is 0.0148 mol.

Now, calculate the mass of AgCl using its molar mass:

• Molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol

$\text{Mass of AgCl} = \text{Moles of AgCl} \times \text{Molar mass of AgCl}$ $\text{Mass of AgCl} = 0.0148 \, \text{mol} \times 143.32 \, \text{g/mol} = 2.1211 \, \text{g}$

So, the theoretical mass of AgCl produced is 2.1211 g.

Step 4: Calculate the percent yield

The actual yield of AgCl is given as 0.770 g. The percent yield is calculated as:

$\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100$ $\text{Percent yield} = \left( \frac{0.770 \, \text{g}}{2.1211 \, \text{g}} \right) \times 100 = 36.29\%$

Thus, the percent yield of AgCl is 36.29%.

Summary of answers:

• Limiting reactant: AgNO₃
• Mass of AgCl produced: 2.1211 g
• Percent yield: 36.29%

Would you like more details on any of these steps? Here are some related questions to deepen your understanding:

1. How would the calculation change if NaCl were the limiting reactant?
2. What factors could lead to a low percent yield in a chemical reaction?
3. How do you calculate the molarity of a solution?
4. Why is it important to know the limiting reactant in a reaction?
5. How can you improve the yield of a reaction in a laboratory setting?

Tip: Always check that your units are consistent when performing calculations in chemistry, especially between mL and L for volume and M for molarity.