Balancing chemical equations and identifying reaction types in a chemistry worksheet.

A worksheet titled "Types of Chemical Reaction Worksheet" with 20 numbered chemical equations to balance and identify the reaction type (synthesis, decomposition, single-displacement, double-displacement, or combustion).

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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 ammonium ions ($\text{NH}_4^+$), so we need 2 $\text{NH}_3$ molecules on the left.
- 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 in $\text{C}_6\text{H}_{12}\text{O}_6$, so we need 6 $\text{CO}_2$ molecules on the right.
- Next, balance hydrogen ($\text{H}$): There are 12 hydrogen atoms in $\text{C}_6\text{H}_{12}\text{O}_6$, so we need 6 $\text{H}_2\text{O}$ molecules on the right.
- Finally, balance oxygen ($\text{O}$): There are 6 oxygen atoms in $\text{C}_6\text{H}_{12}\text{O}_6$ and 18 oxygen atoms in $6\text{CO}_2$ and $6\text{H}_2\text{O}$ (total 24 oxygen atoms on the right). Therefore, we need 6 $\text{O}_2$ molecules on the left.
- 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 in $\text{Pb}_3(\text{PO}_4)_2$, so we need 3 $\text{Pb}$ atoms on the left.
- Next, balance phosphorus ($\text{P}$): There are 2 phosphorus atoms in $\text{Pb}_3(\text{PO}_4)_2$, so we need 2 $\text{H}_3\text{PO}_4$ molecules on the left.
- Balance hydrogen ($\text{H}$): There are 6 hydrogen atoms in $2\text{H}_3\text{PO}_4$, so we need 3 $\text{H}_2$ molecules on the right.
- 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 from $\text{H}_3\text{PO}_4$).

---

#### 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 in $\text{Li}_3\text{N}$, so we need 3 $\text{LiNO}_3$ molecules on the right.
- Next, balance nitrogen ($\text{N}$): There are 4 nitrogen atoms in $(\text{NH}_4)_3\text{N}$, so we need 4 $\text{NH}_4\text{NO}_3$ molecules on the left.
- The balanced equation is:
$$
\text{Li}_3\text{N} + 4\text{NH}_4\text{NO}_3 \rightarrow 3\text{LiNO}_3 + (\text{NH}_4)_3\text{N}
$$

- 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 in $\text{Al(OH)}_3$ and $\text{AlBr}_3$, so no adjustment is needed.
- Balance bromine ($\text{Br}$): There are 3 bromine atoms in $\text{AlBr}_3$, so we need 3 $\text{HBr}$ molecules on the left.
- Balance hydrogen ($\text{H}$): There are 3 hydrogen atoms in $3\text{HBr}$ and 3 hydrogen atoms in $\text{Al(OH)}_3$, so we need 3 $\text{H}_2\text{O}$ molecules on the right.
- 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} + 8\text{O}_2 \rightarrow 5\text{CO}_2 + 6\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 + 2\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:

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

Parent Tip: Review the logic above to help your child master the concept of types of reactions worksheet.

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