Classifying Chemical Reactions worksheet for students to identify and balance chemical equations.

Problem: Classifying Chemical Reactions and Balancing Equations

#### Part A: Classify the following reactions as synthesis, decomposition, single-replacement, double-replacement, or combustion.

1. Reaction 1:
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3
\]
- Analysis: This reaction involves two compounds exchanging ions to form two new compounds.
- Type: Double-replacement

2. Reaction 2:
\[
\text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 2\text{S}
\]
- Analysis: A single compound breaks down into its constituent elements.
- Type: Decomposition

3. Reaction 3:
\[
2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
- Analysis: Two compounds exchange ions to form two new compounds.
- Type: Double-replacement

4. Reaction 4:
\[
\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2
\]
- Analysis: A single element (iron) replaces hydrogen in sulfuric acid.
- Type: Single-replacement

5. Reaction 5:
\[
\text{C}_4\text{H}_{12} + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}
\]
- Analysis: A hydrocarbon reacts with oxygen to produce carbon dioxide and water.
- Type: Combustion

---

#### Part B: Balance the following equations and indicate the type of reaction.

1. Reaction 1:
\[
\text{H}_2\text{S} + \text{O}_2 \rightarrow \text{SO}_2 + \text{H}_2\text{O}
\]
- Balancing:
- Start with sulfur (S): There is 1 S on both sides.
- Balance oxygen (O): There are 2 O atoms in \(\text{O}_2\) and 2 O atoms in \(\text{SO}_2\), but we need 1 more O for \(\text{H}_2\text{O}\). Add a coefficient of 2 to \(\text{H}_2\text{O}\).
- Balance hydrogen (H): There are 2 H atoms in \(\text{H}_2\text{S}\) and 2 H atoms in \(2\text{H}_2\text{O}\). Add a coefficient of 2 to \(\text{H}_2\text{S}\).
- Final balanced equation:
\[
2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O}
\]
- Type: Oxidation-reduction (specifically, oxidation of \(\text{H}_2\text{S}\) by \(\text{O}_2\)).

2. Reaction 2:
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- Balancing:
- Carbon (C): There are 5 C atoms in \(\text{C}_5\text{H}_{10}\text{O}\) and 5 C atoms in \(5\text{CO}_2\).
- Hydrogen (H): There are 10 H atoms in \(\text{C}_5\text{H}_{10}\text{O}\) and 10 H atoms in \(5\text{H}_2\text{O}\).
- Oxygen (O): There are 1 O atom in \(\text{C}_5\text{H}_{10}\text{O}\) and 10 O atoms in \(5\text{CO}_2\) and \(5\text{H}_2\text{O}\), totaling 15 O atoms. Therefore, add a coefficient of 7 to \(\text{O}_2\).
- Final balanced equation:
\[
\text{C}_5\text{H}_{10}\text{O} + 7\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
- Type: Combustion

3. Reaction 3:
\[
\text{Al} + \text{NiBr}_2 \rightarrow \text{AlBr}_3 + \text{Ni}
\]
- Balancing:
- Aluminum (Al): There is 1 Al on the left and 1 Al on the right.
- Nickel (Ni): There is 1 Ni on the left and 1 Ni on the right.
- Bromine (Br): There are 2 Br atoms in \(\text{NiBr}_2\) and 3 Br atoms in \(\text{AlBr}_3\). To balance Br, add a coefficient of 2 to \(\text{AlBr}_3\) and 2 to \(\text{NiBr}_2\). Then, add a coefficient of 2 to \(\text{Al}\).
- Final balanced equation:
\[
2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni}
\]
- Type: Single-replacement

4. Reaction 4:
\[
\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3
\]
- Balancing:
- Aluminum (Al): There are 2 Al atoms in \(\text{Al}_2\text{O}_3\). Add a coefficient of 2 to \(\text{Al}\).
- Oxygen (O): There are 3 O atoms in \(\text{Al}_2\text{O}_3\) and 2 O atoms in \(\text{O}_2\). To balance O, add a coefficient of \( \frac{3}{2} \) to \(\text{O}_2\). Multiply through by 2 to eliminate the fraction.
- Final balanced equation:
\[
4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
\]
- Type: Synthesis

5. Reaction 5:
\[
\text{H}_2\text{O}_2 \rightarrow \text{H}_2 + \text{O}_2
\]
- Balancing:
- Hydrogen (H): There are 2 H atoms in \(\text{H}_2\text{O}_2\) and 2 H atoms in \(\text{H}_2\).
- Oxygen (O): There are 2 O atoms in \(\text{H}_2\text{O}_2\) and 2 O atoms in \(\text{O}_2\).
- The equation is already balanced.
- Final balanced equation:
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2 + \text{O}_2
\]
- Type: Decomposition

---

Final Answers:

#### Part A:
1. Double-replacement
2. Decomposition
3. Double-replacement
4. Single-replacement
5. Combustion

#### Part B:
1. \(2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O}\) (Oxidation-reduction)
2. \(\text{C}_5\text{H}_{10}\text{O} + 7\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}\) (Combustion)
3. \(2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni}\) (Single-replacement)
4. \(4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3\) (Synthesis)
5. \(2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2 + \text{O}_2\) (Decomposition)

Boxed Final Answer:
\[
\boxed{
\begin{array}{ll}
\text{Part A:} & \text{1. Double-replacement, 2. Decomposition, 3. Double-replacement, 4. Single-replacement, 5. Combustion} \\
\text{Part B:} & \text{1. } 2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O} \text{ (Oxidation-reduction)} \\
& \text{2. } \text{C}_5\text{H}_{10}\text{O} + 7\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O} \text{ (Combustion)} \\
& \text{3. } 2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni} \text{ (Single-replacement)} \\
& \text{4. } 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \text{ (Synthesis)} \\
& \text{5. } 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2 + \text{O}_2 \text{ (Decomposition)}
\end{array}
}
\]