11. Matching Type: Balance the given equations in Column I, then choose the letter under Column 2 that corresponds to the sum of the coefficients of each balanced equation. Identify the type of chemical reaction by choosing the letter under Column 3.(15 points) Example: 1-2. $$ 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow \mathbf{2} \mathrm{H}_{2} \mathrm{O} $$ $$ 2+1+2=5 $$ A. 3 B. 4 C. 5 D. 6 Synthesis Answers: $$ \begin{array}{l} \text { 1. } \mathrm{C} \text { (2 points) } \ \text { 2. A (1 point) } \ \text { Column } 1\end{array} $$ $$ \begin{array}{l} \text { 1. } \mathrm{C} \text { (2 points) } \ \text { 2. A (1 point) } \ \text { Column } 1\end{array} $$ $$ \begin{array}{l} \text { 1. } \mathrm{C} \text { (2 points) } \ \text { 2. A (1 point) } \ \text { Column } 1\end{array} $$ $$ \begin{array}{l} \text { 1. } \mathrm{C} \text { (2 points) } \ \text { 2. A (1 point) } \ \text { Column } 1\end{array} $$ 22-23. $$ \mathrm{MnO}_{2}+\mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} $$ 24-25. $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{HCl} \rightarrow \mathrm{FeCl}_{3}+\mathrm{H}_{2} \mathrm{O} $$ $$ 26-27 . \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \rightarrow \mathrm{Nal}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} $$ $$ 28-29 . \mathrm{Na}+\mathrm{KNO}_{3} \rightarrow \mathrm{~K}_{2} \mathrm{O}+\mathrm{Na}_{2} \mathrm{O}+\mathrm{N}_{2} $$ Column 2 A. 6 B. 7 C. 8 D. 9 A. 6 B. 7 C. 8 D. 9 A. 12 B. 13 C. 14 D. 15 A. 6 B. 7 C. 8 D. 9 A. 10 B. 16 C. 19 D. 20 Column 3 A. Synthesis B. Decomposition C. Single Replacement D. Double Replacement

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To solve the problem, we need to balance each of the given chemical equations, calculate the sum of the coefficients for each balanced equation, and identify the type of chemical reaction. Let's go through each equation step by step.

### 1. Balancing the Equations

#### Equation 1: 
$$
\mathrm{MnO}_{2} + \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2} + \mathrm{Cl}_{2} + \mathrm{H}_{2}\mathrm{O}
$$

**Balancing Steps:**
- Start with Mn: 1 Mn on both sides.
- Balance Cl: 2 Cl in $$\mathrm{MnCl}_{2}$$ and 2 Cl in $$\mathrm{Cl}_{2}$$ gives a total of 4 Cl on the right. Therefore, we need 4 HCl on the left.
- Balance O: 2 O in $$\mathrm{MnO}_{2}$$ means we need 2 H2O on the right.

**Balanced Equation:**
$$
\mathrm{MnO}_{2} + 4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2} + 2 \mathrm{Cl}_{2} + 2 \mathrm{H}_{2}\mathrm{O}
$$

**Sum of Coefficients:**
$$
1 + 4 + 1 + 2 + 2 = 10
$$

**Type of Reaction:** 
This is a **Single Replacement** reaction.

---

#### Equation 2: 
$$
\mathrm{Fe}_{2}\mathrm{O}_{3} + \mathrm{HCl} \rightarrow \mathrm{FeCl}_{3} + \mathrm{H}_{2}\mathrm{O}
$$

**Balancing Steps:**
- Start with Fe: 2 Fe on the left means we need 2 $$\mathrm{FeCl}_{3}$$ on the right.
- Balance Cl: 2 $$\mathrm{FeCl}_{3}$$ gives 6 Cl, so we need 6 HCl on the left.
- Balance O: 3 O in $$\mathrm{Fe}_{2}\mathrm{O}_{3}$$ means we need 3 H2O on the right.

**Balanced Equation:**
$$
\mathrm{Fe}_{2}\mathrm{O}_{3} + 6 \mathrm{HCl} \rightarrow 2 \mathrm{FeCl}_{3} + 3 \mathrm{H}_{2}\mathrm{O}
$$

**Sum of Coefficients:**
$$
1 + 6 + 2 + 3 = 12
$$

**Type of Reaction:** 
This is a **Double Replacement** reaction.

---

#### Equation 3: 
$$
\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} + \mathrm{I}_{2} \rightarrow \mathrm{NaI} + \mathrm{Na}_{2}\mathrm{S}_{4}\mathrm{O}_{6}
$$

**Balancing Steps:**
- Start with Na: 2 Na on the left means we need 2 NaI on the right.
- Balance I: 2 I on the left means we need 1 I2 on the left.
- Balance S and O: The sulfur and oxygen are already balanced.

**Balanced Equation:**
$$
\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} + \mathrm{I}_{2} \rightarrow 2 \mathrm{NaI} + \mathrm{Na}_{2}\mathrm{S}_{4}\mathrm{O}_{6}
$$

**Sum of Coefficients:**
$$
1 + 1 + 2 + 1 = 5
$$

**Type of Reaction:** 
This is a **Redox Reaction** (not listed in your options, but it is not a synthesis, decomposition, or replacement).

---

#### Equation 4: 
$$
\mathrm{Na} + \mathrm{KNO}_{3} \rightarrow \mathrm{K}_{2}\mathrm{O} + \mathrm{Na}_{2}\mathrm{O} + \mathrm{N}_{2}
$$

**Balancing Steps:**
- Start with Na: 2 Na on the right means we need 2 Na on the left.
- Balance K: 2 K on the right means we need 2 KNO3 on the left.
- Balance N: 2 N on the right means we need 2 KNO3 on the left.
- Balance O: 6 O on the left means we need 3 H2O on the right.

**Balanced Equation:**
$$
2 \mathrm{Na} + 2 \mathrm{KNO}_{3} \rightarrow \mathrm{K}_{2}\mathrm{O} + \mathrm{Na}_{2}\mathrm{O} + \mathrm{N}_{2}
$$

**Sum of Coefficients:**
$$
2 + 2 + 1 + 2 + 1 = 8
$$

**Type of Reaction:** 
This is a **Single Replacement** reaction.

---

### Summary of Results

1. **Equation 1:** Sum = 10, Type = C (Single Replacement)
2. **Equation 2:** Sum = 12, Type = D (Double Replacement)
3. **Equation 3:** Sum = 5, Type = C (Redox)
4. **Equation 4:** Sum = 8, Type = C (Single Replacement)

### Final Answers
- **1. C (10 points)**
- **2. D (12 points)**
- **3. C (5 points)**
- **4. C (8 points)**

Please note that the types of reactions for the third equation may not match the provided options, as it is a redox reaction.
1. C (10 points) 2. D (12 points) 3. C (5 points) 4. C (8 points)

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