Balance and classify chemical reactions on this educational worksheet.

Problem: Balancing Chemical Reactions and Identifying Reaction Types

The task involves balancing chemical equations and identifying the type of reaction (synthesis, decomposition, single-displacement, double-displacement, or combustion). Below is a detailed solution for each problem.

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#### 1. Balance the reaction and identify the reaction type:
$$
\text{NaBr} + \text{Ca(OH)}_2 \rightarrow \text{CaBr}_2 + \text{NaOH}
$$

- Balancing:
- Start with calcium bromide (\(\text{CaBr}_2\)): There are 2 bromine atoms on the right but only 1 on the left. Add a coefficient of 2 to \(\text{NaBr}\).
- Now, the equation is:
$$
2\text{NaBr} + \text{Ca(OH)}_2 \rightarrow \text{CaBr}_2 + \text{NaOH}
$$
- Balance sodium (\(\text{Na}\)): There are 2 sodium atoms on the left but only 1 on the right. Add a coefficient of 2 to \(\text{NaOH}\).
- The balanced equation is:
$$
2\text{NaBr} + \text{Ca(OH)}_2 \rightarrow \text{CaBr}_2 + 2\text{NaOH}
$$

- Reaction Type: Double-displacement (two compounds exchange ions).

---

#### 2. Balance the reaction and identify the reaction type:
$$
\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4
$$

- Balancing:
- On the right side, there are 2 nitrogen atoms, 8 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.
- To balance nitrogen, add a coefficient of 2 to \(\text{NH}_3\).
- To balance hydrogen, add a coefficient of 1 to \(\text{H}_2\text{SO}_4\) (it is already balanced).
- The balanced equation is:
$$
2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4
$$

- Reaction Type: Synthesis (two substances combine to form one compound).

---

#### 3. Balance the reaction and identify the reaction type:
$$
\text{C}_6\text{H}_{12}\text{O}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
$$

- Balancing:
- Start with carbon (\(\text{C}\)): There are 6 carbon atoms on the left but only 1 on the right. Add a coefficient of 6 to \(\text{CO}_2\).
- Balance hydrogen (\(\text{H}\)): There are 12 hydrogen atoms on the left but only 2 on the right. Add a coefficient of 6 to \(\text{H}_2\text{O}\).
- Balance oxygen (\(\text{O}\)): There are 6 oxygen atoms in \(\text{C}_6\text{H}_{12}\text{O}_6\) and 6 more in \(\text{O}_2\) (total 12 on the left). On the right, there are \(6 \times 2 = 12\) oxygen atoms in \(\text{CO}_2\) and \(6 \times 1 = 6\) oxygen atoms in \(\text{H}_2\text{O}\) (total 18). Add a coefficient of 6 to \(\text{O}_2\) to balance.
- The balanced equation is:
$$
\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}
$$

- Reaction Type: Combustion (a substance reacts with oxygen to produce carbon dioxide and water).

---

#### 4. Balance the reaction and identify the reaction type:
$$
\text{Pb} + \text{H}_3\text{PO}_4 \rightarrow \text{H}_2 + \text{Pb}_3(\text{PO}_4)_2
$$

- Balancing:
- Start with lead (\(\text{Pb}\)): There are 3 lead atoms on the right but only 1 on the left. Add a coefficient of 3 to \(\text{Pb}\).
- Balance hydrogen (\(\text{H}\)): There are 3 hydrogen atoms in \(\text{H}_3\text{PO}_4\) but 2 on the right. Add a coefficient of 3 to \(\text{H}_2\).
- Balance phosphorus (\(\text{P}\)) and oxygen (\(\text{O}\)): The coefficients are already balanced.
- The balanced equation is:
$$
3\text{Pb} + 2\text{H}_3\text{PO}_4 \rightarrow 3\text{H}_2 + \text{Pb}_3(\text{PO}_4)_2
$$

- Reaction Type: Single-displacement (lead displaces hydrogen).

---

#### 5. Balance the reaction and identify the reaction type:
$$
\text{Li}_3\text{N} + \text{NH}_4\text{NO}_3 \rightarrow \text{LiNO}_3 + (\text{NH}_4)_3\text{N}
$$

- Balancing:
- Start with lithium (\(\text{Li}\)): There are 3 lithium atoms on the left but only 1 on the right. Add a coefficient of 3 to \(\text{LiNO}_3\).
- Balance nitrogen (\(\text{N}\)): There are 4 nitrogen atoms in \(\text{NH}_4\text{NO}_3\) but only 1 on the left. Add a coefficient of 4 to \(\text{Li}_3\text{N}\).
- Balance hydrogen (\(\text{H}\)): There are 4 hydrogen atoms in \(\text{NH}_4\text{NO}_3\) but 12 on the right. The coefficients are already balanced.
- The balanced equation is:
$$
4\text{Li}_3\text{N} + 3\text{NH}_4\text{NO}_3 \rightarrow 12\text{LiNO}_3 + (\text{NH}_4)_3\text{N}
$$

- Reaction Type: Double-displacement (ions are exchanged between two compounds).

---

#### 6. Balance the reaction and identify the reaction type:
$$
\text{HBr} + \text{Al(OH)}_3 \rightarrow \text{H}_2\text{O} + \text{AlBr}_3
$$

- Balancing:
- Start with aluminum (\(\text{Al}\)): There is 1 aluminum atom on the left but 1 on the right. It is balanced.
- Balance bromine (\(\text{Br}\)): There is 1 bromine atom on the left but 3 on the right. Add a coefficient of 3 to \(\text{HBr}\).
- Balance hydrogen (\(\text{H}\)): There are 3 hydrogen atoms in \(\text{Al(OH)}_3\) but 2 on the right. Add a coefficient of 3 to \(\text{H}_2\text{O}\).
- The balanced equation is:
$$
3\text{HBr} + \text{Al(OH)}_3 \rightarrow 3\text{H}_2\text{O} + \text{AlBr}_3
$$

- Reaction Type: Double-displacement (ions are exchanged between two compounds).

---

#### 7–20. Identify the reaction type for each given reaction:

1. \( \text{Na}_3\text{PO}_4 + 3\text{KOH} \rightarrow 3\text{NaOH} + \text{K}_3\text{PO}_4 \)
- Type: Double-displacement

2. \( \text{MgCl}_2 + \text{Li}_2\text{CO}_3 \rightarrow \text{MgCO}_3 + 2\text{LiCl} \)
- Type: Double-displacement

3. \( \text{C}_9\text{H}_{12} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O} \)
- Type: Combustion

4. \( \text{Pb} + \text{FeSO}_4 \rightarrow \text{PbSO}_4 + \text{Fe} \)
- Type: Single-displacement

5. \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)
- Type: Decomposition

6. \( \text{P}_4 + 3\text{O}_2 \rightarrow 2\text{P}_2\text{O}_3 \)
- Type: Synthesis

7. \( 2\text{RbNO}_3 + \text{BeF}_2 \rightarrow \text{Be(NO}_3)_2 + 2\text{RbF} \)
- Type: Double-displacement

8. \( 2\text{AgNO}_3 + \text{Cu} \rightarrow \text{Cu(NO}_3)_2 + 2\text{Ag} \)
- Type: Single-displacement

9. \( \text{C}_5\text{H}_{12} + 4\text{O}_2 \rightarrow 3\text{CO}_2 + 3\text{H}_2\text{O} \)
- Type: Combustion

10. \( 2\text{C}_8\text{H}_{18} + \text{Fe} \rightarrow \text{Fe(C}_8\text{H}_{18})_2 \)
- Type: Synthesis

11. \( \text{SeCl}_4 + \text{O}_2 \rightarrow \text{SeO}_2 + 3\text{Cl}_2 \)
- Type: Decomposition

12. \( 2\text{MgI}_2 + \text{Mn(SO}_4)_2 \rightarrow 2\text{MgSO}_4 + \text{MnI}_4 \)
- Type: Double-displacement

13. \( \text{O}_3 \rightarrow 3\text{O}_2 \)
- Type: Decomposition

14. \( 2\text{NO}_2 \rightarrow 2\text{O}_2 + \text{N}_2 \)
- Type: Decomposition

---

Final Answer:

$$
\boxed{
\begin{array}{ll}
1. & 2\text{NaBr} + \text{Ca(OH)}_2 \rightarrow \text{CaBr}_2 + 2\text{NaOH}, \text{Double-displacement} \\
2. & 2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4, \text{Synthesis} \\
3. & \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}, \text{Combustion} \\
4. & 3\text{Pb} + 2\text{H}_3\text{PO}_4 \rightarrow 3\text{H}_2 + \text{Pb}_3(\text{PO}_4)_2, \text{Single-displacement} \\
5. & 4\text{Li}_3\text{N} + 3\text{NH}_4\text{NO}_3 \rightarrow 12\text{LiNO}_3 + (\text{NH}_4)_3\text{N}, \text{Double-displacement} \\
6. & 3\text{HBr} + \text{Al(OH)}_3 \rightarrow 3\text{H}_2\text{O} + \text{AlBr}_3, \text{Double-displacement} \\
7. & \text{Double-displacement} \\
8. & \text{Double-displacement} \\
9. & \text{Combustion} \\
10. & \text{Single-displacement} \\
11. & \text{Decomposition} \\
12. & \text{Double-displacement} \\
13. & \text{Decomposition} \\
14. & \text{Decomposition} \\
\end{array}
}
$$