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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.

A 74.0 mL solution of 0.200 M of AgNO₃ reacts with 195.0 mL solution of 0.351 M of NaCl to produce silver chloride (AgCl) and sodium nitrate (NaNO₃) according to the balanced equation: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). Determine the limiting reactant for the given reaction. Calculate the mass of AgCl produced in the reaction and the percent yield if 0.770 g of AgCl are isolated after carrying out the reaction.

Learn how to determine the limiting reactant and calculate the mass of silver chloride produced in a reaction between silver nitrate and sodium chloride. Understand key stoichiometric calculations, including moles, molarity, and percent yield. Suitable for Grades 10-12 chemistry students.

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