Free Balancing Chemical Equations Worksheet Answer Key - Free Printable

Task: Balance the chemical equations and classify each reaction as synthesis, decomposition, single-replacement, or double-replacement.

#### Step-by-Step Solution:

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1. \( \_\_\_ \text{Sb} + \_\_\_ \text{Cl}_2 \rightarrow \_\_\_ \text{SbCl}_3 \)

- Balancing:
- Sb: 1 atom on the left, 1 atom on the right.
- Cl: 2 atoms on the left, 3 atoms on the right.
- To balance Cl, multiply SbCl₃ by 2 and Cl₂ by 3:
\[
2 \text{Sb} + 3 \text{Cl}_2 \rightarrow 2 \text{SbCl}_3
\]

- Classification: This is a synthesis reaction because two elements combine to form a compound.

Answer:
\[
2 \text{Sb} + 3 \text{Cl}_2 \rightarrow 2 \text{SbCl}_3 \quad \text{(Synthesis)}
\]

---

2. \( \_\_\_ \text{Mg} + \_\_\_ \text{O}_2 \rightarrow \_\_\_ \text{MgO} \)

- Balancing:
- Mg: 1 atom on the left, 1 atom on the right.
- O: 2 atoms on the left, 1 atom on the right.
- To balance O, multiply MgO by 2:
\[
2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO}
\]

- Classification: This is a synthesis reaction because two elements combine to form a compound.

Answer:
\[
2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \quad \text{(Synthesis)}
\]

---

3. \( \_\_\_ \text{CaCl}_2 \rightarrow \_\_\_ \text{Ca} + \_\_\_ \text{Cl}_2 \)

- Balancing:
- Ca: 1 atom on the left, 1 atom on the right.
- Cl: 2 atoms on the left, 2 atoms on the right.
- The equation is already balanced.

- Classification: This is a decomposition reaction because a compound breaks down into its elements.

Answer:
\[
\text{CaCl}_2 \rightarrow \text{Ca} + \text{Cl}_2 \quad \text{(Decomposition)}
\]

---

4. \( \_\_\_ \text{NaClO}_3 \rightarrow \_\_\_ \text{NaCl} + \_\_\_ \text{O}_2 \)

- Balancing:
- Na: 1 atom on the left, 1 atom on the right.
- Cl: 1 atom on the left, 1 atom on the right.
- O: 3 atoms on the left, 2 atoms on the right.
- To balance O, multiply NaClO₃ by 2 and O₂ by 3:
\[
2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2
\]

- Classification: This is a decomposition reaction because a compound breaks down into simpler substances.

Answer:
\[
2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2 \quad \text{(Decomposition)}
\]

---

5. \( \_\_\_ \text{Fe} + \_\_\_ \text{HCl} \rightarrow \_\_\_ \text{FeCl}_2 + \_\_\_ \text{H}_2 \)

- Balancing:
- Fe: 1 atom on the left, 1 atom on the right.
- H: 1 atom on the left, 2 atoms on the right.
- Cl: 1 atom on the left, 2 atoms on the right.
- To balance H and Cl, multiply HCl by 2:
\[
\text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2
\]

- Classification: This is a single-replacement reaction because Fe replaces H in HCl.

Answer:
\[
\text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \quad \text{(Single-replacement)}
\]

---

6. \( \_\_\_ \text{CuO} + \_\_\_ \text{H}_2 \rightarrow \_\_\_ \text{Cu} + \_\_\_ \text{H}_2\text{O} \)

- Balancing:
- Cu: 1 atom on the left, 1 atom on the right.
- O: 1 atom on the left, 1 atom on the right.
- H: 2 atoms on the left, 2 atoms on the right.
- The equation is already balanced.

- Classification: This is a single-replacement reaction because H₂ replaces Cu in CuO.

Answer:
\[
\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \quad \text{(Single-replacement)}
\]

---

7. \( \_\_\_ \text{Al} + \_\_\_ \text{H}_2\text{SO}_4 \rightarrow \_\_\_ \text{Al}_2(\text{SO}_4)_3 + \_\_\_ \text{H}_2 \)

- Balancing:
- Al: 1 atom on the left, 2 atoms on the right.
- H: 2 atoms on the left, 2 atoms on the right.
- S: 1 atom on the left, 3 atoms on the right.
- O: 4 atoms on the left, 12 atoms on the right.
- To balance Al, multiply Al by 2. To balance S and O, multiply H₂SO₄ by 3:
\[
2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2
\]

- Classification: This is a single-replacement reaction because Al replaces H in H₂SO₄.

Answer:
\[
2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2 \quad \text{(Single-replacement)}
\]

---

8. \( \_\_\_ \text{MgBr}_2 + \_\_\_ \text{Cl}_2 \rightarrow \_\_\_ \text{MgCl}_2 + \_\_\_ \text{Br}_2 \)

- Balancing:
- Mg: 1 atom on the left, 1 atom on the right.
- Br: 2 atoms on the left, 2 atoms on the right.
- Cl: 2 atoms on the left, 2 atoms on the right.
- The equation is already balanced.

- Classification: This is a double-replacement reaction because the cations (Mg²⁺ and Br⁻) switch places.

Answer:
\[
\text{MgBr}_2 + \text{Cl}_2 \rightarrow \text{MgCl}_2 + \text{Br}_2 \quad \text{(Double-replacement)}
\]

---

9. \( \_\_\_ \text{SnO}_2 + \_\_\_ \text{C} \rightarrow \_\_\_ \text{Sn} + \_\_\_ \text{CO} \)

- Balancing:
- Sn: 1 atom on the left, 1 atom on the right.
- O: 2 atoms on the left, 1 atom on the right.
- C: 1 atom on the left, 1 atom on the right.
- To balance O, multiply CO by 2:
\[
\text{SnO}_2 + 2 \text{C} \rightarrow \text{Sn} + 2 \text{CO}
\]

- Classification: This is a single-replacement reaction because C replaces O in SnO₂.

Answer:
\[
\text{SnO}_2 + 2 \text{C} \rightarrow \text{Sn} + 2 \text{CO} \quad \text{(Single-replacement)}
\]

---

10. \( \_\_\_ \text{Pb(NO}_3\text{)}_2 + \_\_\_ \text{H}_2\text{S} \rightarrow \_\_\_ \text{PbS} + \_\_\_ \text{HNO}_3 \)

- Balancing:
- Pb: 1 atom on the left, 1 atom on the right.
- N: 2 atoms on the left, 2 atoms on the right.
- O: 6 atoms on the left, 3 atoms on the right.
- H: 2 atoms on the left, 2 atoms on the right.
- S: 1 atom on the left, 1 atom on the right.
- To balance O, multiply HNO₃ by 2:
\[
\text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2 \text{HNO}_3
\]

- Classification: This is a double-replacement reaction because the ions switch places.

Answer:
\[
\text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2 \text{HNO}_3 \quad \text{(Double-replacement)}
\]

---

11. \( \_\_\_ \text{HgO} \rightarrow \_\_\_ \text{Hg} + \_\_\_ \text{O}_2 \)

- Balancing:
- Hg: 1 atom on the left, 1 atom on the right.
- O: 1 atom on the left, 2 atoms on the right.
- To balance O, multiply HgO by 2 and O₂ by 1:
\[
2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2
\]

- Classification: This is a decomposition reaction because a compound breaks down into its elements.

Answer:
\[
2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2 \quad \text{(Decomposition)}
\]

---

12. \( \_\_\_ \text{KClO}_3 \rightarrow \_\_\_ \text{KCl} + \_\_\_ \text{O}_2 \)

- Balancing:
- K: 1 atom on the left, 1 atom on the right.
- Cl: 1 atom on the left, 1 atom on the right.
- O: 3 atoms on the left, 2 atoms on the right.
- To balance O, multiply KClO₃ by 2 and O₂ by 3:
\[
2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2
\]

- Classification: This is a decomposition reaction because a compound breaks down into simpler substances.

Answer:
\[
2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \quad \text{(Decomposition)}
\]

---

13. \( \_\_\_ \text{N}_2 + \_\_\_ \text{H}_2 \rightarrow \_\_\_ \text{NH}_3 \)

- Balancing:
- N: 2 atoms on the left, 1 atom on the right.
- H: 2 atoms on the left, 3 atoms on the right.
- To balance N, multiply NH₃ by 2. To balance H, multiply H₂ by 3:
\[
\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3
\]

- Classification: This is a synthesis reaction because two elements combine to form a compound.

Answer:
\[
\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \quad \text{(Synthesis)}
\]

---

14. \( \_\_\_ \text{NaBr} + \_\_\_ \text{Cl}_2 \rightarrow \_\_\_ \text{NaCl} + \_\_\_ \text{Br}_2 \)

- Balancing:
- Na: 1 atom on the left, 1 atom on the right.
- Br: 1 atom on the left, 2 atoms on the right.
- Cl: 2 atoms on the left, 1 atom on the right.
- To balance Br and Cl, multiply NaBr by 2 and NaCl by 2:
\[
2 \text{NaBr} + \text{Cl}_2 \rightarrow 2 \text{NaCl} + \text{Br}_2
\]

- Classification: This is a single-replacement reaction because Cl₂ replaces Br in NaBr.

Answer:
\[
2 \text{NaBr} + \text{Cl}_2 \rightarrow 2 \text{NaCl} + \text{Br}_2 \quad \text{(Single-replacement)}
\]

---

15. \( \_\_\_ \text{Zn} + \_\_\_ \text{AgNO}_3 \rightarrow \_\_\_ \text{Zn(NO}_3\text{)}_2 + \_\_\_ \text{Ag} \)

- Balancing:
- Zn: 1 atom on the left, 1 atom on the right.
- Ag: 1 atom on the left, 1 atom on the right.
- N: 1 atom on the left, 2 atoms on the right.
- O: 3 atoms on the left, 6 atoms on the right.
- To balance N and O, multiply AgNO₃ by 2:
\[
\text{Zn} + 2 \text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + 2 \text{Ag}
\]

- Classification: This is a single-replacement reaction because Zn replaces Ag in AgNO₃.

Answer:
\[
\text{Zn} + 2 \text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + 2 \text{Ag} \quad \text{(Single-replacement)}
\]

---

16. \( \_\_\_ \text{Sn} + \_\_\_ \text{Cl}_2 \rightarrow \_\_\_ \text{SnCl}_4 \)

- Balancing:
- Sn: 1 atom on the left, 1 atom on the right.
- Cl: 2 atoms on the left, 4 atoms on the right.
- To balance Cl, multiply Cl₂ by 2:
\[
\text{Sn} + 2 \text{Cl}_2 \rightarrow \text{SnCl}_4
\]

- Classification: This is a synthesis reaction because an element reacts with a compound to form a new compound.

Answer:
\[
\text{Sn} + 2 \text{Cl}_2 \rightarrow \text{SnCl}_4 \quad \text{(Synthesis)}
\]

---

17. \( \_\_\_ \text{Ba(OH)}_2 \rightarrow \_\_\_ \text{BaO} + \_\_\_ \text{H}_2\text{O} \)

- Balancing:
- Ba: 1 atom on the left, 1 atom on the right.
- O: 2 atoms on the left, 1 atom on the right.
- H: 2 atoms on the left, 2 atoms on the right.
- To balance O, multiply BaO by 1 and H₂O by 1:
\[
\text{Ba(OH)}_2 \rightarrow \text{BaO} + \text{H}_2\text{O}
\]

- Classification: This is a decomposition reaction because a compound breaks down into simpler substances.

Answer:
\[
\text{Ba(OH)}_2 \rightarrow \text{BaO} + \text{H}_2\text{O} \quad \text{(Decomposition)}
\]

---

Final Answer:

\[
\boxed{
\begin{aligned}
1. & \ 2 \text{Sb} + 3 \text{Cl}_2 \rightarrow 2 \text{SbCl}_3 \quad \text{(Synthesis)} \\
2. & \ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \quad \text{(Synthesis)} \\
3. & \ \text{CaCl}_2 \rightarrow \text{Ca} + \text{Cl}_2 \quad \text{(Decomposition)} \\
4. & \ 2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2 \quad \text{(Decomposition)} \\
5. & \ \text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \quad \text{(Single-replacement)} \\
6. & \ \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \quad \text{(Single-replacement)} \\
7. & \ 2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2 \quad \text{(Single-replacement)} \\
8. & \ \text{MgBr}_2 + \text{Cl}_2 \rightarrow \text{MgCl}_2 + \text{Br}_2 \quad \text{(Double-replacement)} \\
9. & \ \text{SnO}_2 + 2 \text{C} \rightarrow \text{Sn} + 2 \text{CO} \quad \text{(Single-replacement)} \\
10. & \ \text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2 \text{HNO}_3 \quad \text{(Double-replacement)} \\
11. & \ 2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2 \quad \text{(Decomposition)} \\
12. & \ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \quad \text{(Decomposition)} \\
13. & \ \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \quad \text{(Synthesis)} \\
14. & \ 2 \text{NaBr} + \text{Cl}_2 \rightarrow 2 \text{NaCl} + \text{Br}_2 \quad \text{(Single-replacement)} \\
15. & \ \text{Zn} + 2 \text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + 2 \text{Ag} \quad \text{(Single-replacement)} \\
16. & \ \text{Sn} + 2 \text{Cl}_2 \rightarrow \text{SnCl}_4 \quad \text{(Synthesis)} \\
17. & \ \text{Ba(OH)}_2 \rightarrow \text{BaO} + \text{H}_2\text{O} \quad \text{(Decomposition)}
\end{aligned}
}
\]