Key to balancing chemical equations and classifying reaction types.

Problem: Balancing Chemical Equations and Classifying Reactions

The task involves balancing each chemical equation and classifying the reaction as synthesis, decomposition, single-replacement, or double-replacement. Below is a detailed explanation of how to solve each part.

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#### 1. Balance the Equation
To balance a chemical equation, ensure that the number of atoms of each element is the same on both sides of the equation. This is done by adjusting the coefficients in front of the chemical formulas.

#### 2. Classify the Reaction
- Synthesis: A reaction where two or more substances combine to form a single product.
- Decomposition: A reaction where a single compound breaks down into two or more simpler substances.
- Single-Replacement: A reaction where one element replaces another element in a compound.
- Double-Replacement: A reaction where the ions of two compounds exchange places to form two new compounds.

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Solutions and Explanations

#### 1. \( \_ \_ \text{Sb} + \_ \_ \text{Cl}_2 \rightarrow \_ \_ \text{SbCl}_3 \)
- Balancing:
- Start with Sb: There is 1 Sb on the left and 1 Sb on the right, so no adjustment needed.
- For Cl: There are 2 Cl on the left and 3 Cl on the right. To balance, multiply SbCl₃ by 2 and Cl₂ by 3.
- Balanced equation: \( 2 \text{Sb} + 3 \text{Cl}_2 \rightarrow 2 \text{SbCl}_3 \)
- Classification: Synthesis (two elements combine to form a compound).
- Answer: \( 2 \text{Sb} + 3 \text{Cl}_2 \rightarrow 2 \text{SbCl}_3 \) (synthesis)

#### 2. \( \_ \_ \text{Mg} + \_ \_ \text{O}_2 \rightarrow \_ \_ \text{MgO} \)
- Balancing:
- Mg: 1 Mg on the left, 1 Mg on the right.
- O: 2 O on the left, 1 O on the right. Multiply MgO by 2.
- Balanced equation: \( 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \)
- Classification: Synthesis (elements combine to form a compound).
- Answer: \( 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \) (synthesis)

#### 3. \( \_ \_ \text{CaCl}_2 \rightarrow \_ \_ \text{Ca} + \_ \_ \text{Cl}_2 \)
- Balancing:
- Ca: 1 Ca on the left, 1 Ca on the right.
- Cl: 2 Cl on the left, 2 Cl on the right.
- Balanced equation: \( \text{CaCl}_2 \rightarrow \text{Ca} + \text{Cl}_2 \)
- Classification: Decomposition (a compound breaks down into elements).
- Answer: \( \text{CaCl}_2 \rightarrow \text{Ca} + \text{Cl}_2 \) (decomposition)

#### 4. \( \_ \_ \text{NaClO}_3 \rightarrow \_ \_ \text{NaCl} + \_ \_ \text{O}_2 \)
- Balancing:
- Na: 1 Na on the left, 1 Na on the right.
- Cl: 1 Cl on the left, 1 Cl on the right.
- O: 3 O on the left, 2 O on the right. Multiply O₂ by 3/2 (or adjust to whole numbers by multiplying everything by 2).
- Balanced equation: \( 2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2 \)
- Classification: Decomposition (a compound breaks down into simpler substances).
- Answer: \( 2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2 \) (decomposition)

#### 5. \( \_ \_ \text{Fe} + \_ \_ \text{HCl} \rightarrow \_ \_ \text{FeCl}_2 + \_ \_ \text{H}_2 \)
- Balancing:
- Fe: 1 Fe on the left, 1 Fe on the right.
- H: 2 H on the left, 2 H on the right.
- Cl: 2 Cl on the left, 2 Cl on the right.
- Balanced equation: \( \text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \)
- Classification: Single-replacement (Fe replaces H in HCl).
- Answer: \( \text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2 \) (single replacement)

#### 6. \( \_ \_ \text{CuO} + \_ \_ \text{H}_2 \rightarrow \_ \_ \text{Cu} + \_ \_ \text{H}_2\text{O} \)
- Balancing:
- Cu: 1 Cu on the left, 1 Cu on the right.
- O: 1 O on the left, 1 O on the right.
- H: 2 H on the left, 2 H on the right.
- Balanced equation: \( \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \)
- Classification: Single-replacement (H₂ replaces Cu in CuO).
- Answer: \( \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \) (single replacement)

#### 7. \( \_ \_ \text{Al} + \_ \_ \text{H}_2\text{SO}_4 \rightarrow \_ \_ \text{Al}_2(\text{SO}_4)_3 + \_ \_ \text{H}_2 \)
- Balancing:
- Al: 2 Al on the right, so multiply Al by 2.
- H: 6 H on the right, so multiply H₂SO₄ by 3 and H₂ by 3.
- SO₄: 3 SO₄ on the right, so multiply H₂SO₄ by 3.
- Balanced equation: \( 2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2 \)
- Classification: Single-replacement (Al replaces H in H₂SO₄).
- Answer: \( 2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2 \) (single replacement)

#### 8. \( \_ \_ \text{MgBr}_2 + \_ \_ \text{Cl}_2 \rightarrow \_ \_ \text{MgCl}_2 + \_ \_ \text{Br}_2 \)
- Balancing:
- Mg: 1 Mg on the left, 1 Mg on the right.
- Br: 2 Br on the left, 2 Br on the right.
- Cl: 2 Cl on the left, 2 Cl on the right.
- Balanced equation: \( \text{MgBr}_2 + \text{Cl}_2 \rightarrow \text{MgCl}_2 + \text{Br}_2 \)
- Classification: Single-replacement (Cl₂ replaces Br in MgBr₂).
- Answer: \( \text{MgBr}_2 + \text{Cl}_2 \rightarrow \text{MgCl}_2 + \text{Br}_2 \) (single replacement)

#### 9. \( \_ \_ \text{SnO}_2 + \_ \_ \text{C} \rightarrow \_ \_ \text{Sn} + \_ \_ \text{CO} \)
- Balancing:
- Sn: 1 Sn on the left, 1 Sn on the right.
- O: 2 O on the left, 1 O on the right. Multiply CO by 2.
- C: 1 C on the left, 1 C on the right.
- Balanced equation: \( \text{SnO}_2 + 2 \text{C} \rightarrow \text{Sn} + 2 \text{CO} \)
- Classification: Single-replacement (C replaces O in SnO₂).
- Answer: \( \text{SnO}_2 + 2 \text{C} \rightarrow \text{Sn} + 2 \text{CO} \) (single replacement)

#### 10. \( \_ \_ \text{Pb(NO}_3\text{)}_2 + \_ \_ \text{H}_2\text{S} \rightarrow \_ \_ \text{PbS} + \_ \_ \text{HNO}_3 \)
- Balancing:
- Pb: 1 Pb on the left, 1 Pb on the right.
- N: 2 N on the left, 2 N on the right.
- O: 6 O on the left, 3 O on the right. Multiply HNO₃ by 2.
- S: 1 S on the left, 1 S on the right.
- H: 2 H on the left, 2 H on the right.
- Balanced equation: \( \text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2 \text{HNO}_3 \)
- Classification: Double-replacement (ions exchange places).
- Answer: \( \text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2 \text{HNO}_3 \) (double replacement)

#### 11. \( \_ \_ \text{HgO} \rightarrow \_ \_ \text{Hg} + \_ \_ \text{O}_2 \)
- Balancing:
- Hg: 1 Hg on the left, 1 Hg on the right.
- O: 1 O on the left, 2 O on the right. Multiply HgO by 2.
- Balanced equation: \( 2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2 \)
- Classification: Decomposition (a compound breaks down into elements).
- Answer: \( 2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2 \) (decomposition)

#### 12. \( \_ \_ \text{KClO}_3 \rightarrow \_ \_ \text{KCl} + \_ \_ \text{O}_2 \)
- Balancing:
- K: 1 K on the left, 1 K on the right.
- Cl: 1 Cl on the left, 1 Cl on the right.
- O: 3 O on the left, 2 O on the right. Multiply O₂ by 3/2 (or adjust to whole numbers by multiplying everything by 2).
- Balanced equation: \( 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \)
- Classification: Decomposition (a compound breaks down into simpler substances).
- Answer: \( 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \) (decomposition)

#### 13. \( \_ \_ \text{N}_2 + \_ \_ \text{H}_2 \rightarrow \_ \_ \text{NH}_3 \)
- Balancing:
- N: 2 N on the left, 1 N on the right. Multiply NH₃ by 2.
- H: 2 H on the left, 3 H on the right. Multiply H₂ by 3.
- Balanced equation: \( \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \)
- Classification: Synthesis (elements combine to form a compound).
- Answer: \( \text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3 \) (synthesis)

#### 14. \( \_ \_ \text{NaBr} + \_ \_ \text{Cl}_2 \rightarrow \_ \_ \text{NaCl} + \_ \_ \text{Br}_2 \)
- Balancing:
- Na: 1 Na on the left, 1 Na on the right.
- Br: 1 Br on the left, 1 Br on the right.
- Cl: 2 Cl on the left, 1 Cl on the right. Multiply NaCl by 2.
- Balanced equation: \( 2 \text{NaBr} + \text{Cl}_2 \rightarrow 2 \text{NaCl} + \text{Br}_2 \)
- Classification: Single-replacement (Cl₂ replaces Br in NaBr).
- Answer: \( 2 \text{NaBr} + \text{Cl}_2 \rightarrow 2 \text{NaCl} + \text{Br}_2 \) (single replacement)

#### 15. \( \_ \_ \text{Zn} + \_ \_ \text{AgNO}_3 \rightarrow \_ \_ \text{Zn(NO}_3\text{)}_2 + \_ \_ \text{Ag} \)
- Balancing:
- Zn: 1 Zn on the left, 1 Zn on the right.
- Ag: 1 Ag on the right, 1 Ag on the left. Multiply AgNO₃ by 2.
- NO₃: 2 NO₃ on the right, 2 NO₃ on the left.
- Balanced equation: \( \text{Zn} + 2 \text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + 2 \text{Ag} \)
- Classification: Single-replacement (Zn replaces Ag in AgNO₃).
- Answer: \( \text{Zn} + 2 \text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + 2 \text{Ag} \) (single replacement)

#### 16. \( \_ \_ \text{Sn} + \_ \_ \text{Cl}_2 \rightarrow \_ \_ \text{SnCl}_4 \)
- Balancing:
- Sn: 1 Sn on the left, 1 Sn on the right.
- Cl: 2 Cl on the left, 4 Cl on the right. Multiply Cl₂ by 2.
- Balanced equation: \( \text{Sn} + 2 \text{Cl}_2 \rightarrow \text{SnCl}_4 \)
- Classification: Synthesis (elements combine to form a compound).
- Answer: \( \text{Sn} + 2 \text{Cl}_2 \rightarrow \text{SnCl}_4 \) (synthesis)

#### 17. \( \_ \_ \text{Ba(OH)}_2 \rightarrow \_ \_ \text{BaO} + \_ \_ \text{H}_2\text{O} \)
- Balancing:
- Ba: 1 Ba on the left, 1 Ba on the right.
- O: 2 O on the left, 1 O on the right. Multiply BaO by 1 and H₂O by 1.
- H: 2 H on the left, 2 H on the right.
- Balanced equation: \( \text{Ba(OH)}_2 \rightarrow \text{BaO} + \text{H}_2\text{O} \)
- Classification: Decomposition (a compound breaks down into simpler substances).
- Answer: \( \text{Ba(OH)}_2 \rightarrow \text{BaO} + \text{H}_2\text{O} \) (decomposition)

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Final Answer:

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