Chemistry: Balancing Chemical Equations Worksheet - Free Printable

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 is needed for Sb.
- For Cl: There are 2 Cl atoms on the left (in \(\text{Cl}_2\)) and 3 Cl atoms on the right (in \(\text{SbCl}_3\)). To balance Cl, multiply \(\text{Cl}_2\) by 3/2 and \(\text{SbCl}_3\) by 1. However, since coefficients must be whole numbers, multiply everything by 2:
\[
2 \text{Sb} + 3 \text{Cl}_2 \rightarrow 2 \text{SbCl}_3
\]
- Classification: Synthesis (two elements combine to form a compound).
- Final 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 atoms on the left (in \(\text{O}_2\)) and 1 O atom on the right (in \(\text{MgO}\)). Multiply \(\text{MgO}\) by 2:
\[
2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO}
\]
- Classification: Synthesis (two elements combine to form a compound).
- Final 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 and 1 Ca on the right.
- Cl: 2 Cl atoms on the left and 2 Cl atoms on the right (in \(\text{Cl}_2\)).
- The equation is already balanced.
- Classification: Decomposition (a single compound breaks down into elements).
- Final 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 and 1 Na on the right.
- Cl: 1 Cl on the left and 1 Cl on the right.
- O: 3 O atoms on the left and 2 O atoms on the right (in \(\text{O}_2\)). Multiply \(\text{O}_2\) by 3/2 and adjust coefficients:
\[
2 \text{NaClO}_3 \rightarrow 2 \text{NaCl} + 3 \text{O}_2
\]
- Classification: Decomposition (a single compound breaks down into simpler substances).
- Final 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 and 1 Fe on the right.
- H: 2 H atoms on the right (in \(\text{H}_2\)), so multiply \(\text{HCl}\) by 2.
- Cl: 2 Cl atoms on the right (in \(\text{FeCl}_2\)), so multiply \(\text{HCl}\) by 2.
\[
\text{Fe} + 2 \text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2
\]
- Classification: Single-replacement (Fe replaces H in \(\text{HCl}\)).
- Final 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 and 1 Cu on the right.
- O: 1 O atom on the left and 1 O atom on the right (in \(\text{H}_2\text{O}\)).
- H: 2 H atoms on the left and 2 H atoms on the right (in \(\text{H}_2\text{O}\)).
- The equation is already balanced.
- Classification: Single-replacement (H replaces Cu in \(\text{CuO}\)).
- Final 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 atoms on the right (in \(\text{Al}_2(\text{SO}_4)_3\)), so multiply \(\text{Al}\) by 2.
- H: 6 H atoms on the right (in 3 \(\text{H}_2\)), so multiply \(\text{H}_2\text{SO}_4\) by 3.
- S: 3 S atoms on the right (in \(\text{Al}_2(\text{SO}_4)_3\)), so multiply \(\text{H}_2\text{SO}_4\) by 3.
- O: 12 O atoms on the right (in \(\text{Al}_2(\text{SO}_4)_3\)), which matches 3 \(\text{H}_2\text{SO}_4\).
\[
2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2
\]
- Classification: Single-replacement (Al replaces H in \(\text{H}_2\text{SO}_4\)).
- Final 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 and 1 Mg on the right.
- Br: 2 Br atoms on the left and 2 Br atoms on the right (in \(\text{Br}_2\)).
- Cl: 2 Cl atoms on the left (in \(\text{Cl}_2\)) and 2 Cl atoms on the right (in \(\text{MgCl}_2\)).
- The equation is already balanced.
- Classification: Single-replacement (Cl replaces Br in \(\text{MgBr}_2\)).
- Final 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 and 1 Sn on the right.
- O: 2 O atoms on the left and 1 O atom on the right (in \(\text{CO}\)). Multiply \(\text{CO}\) by 2.
- C: 1 C on the left and 1 C on the right (in \(\text{CO}\)).
\[
\text{SnO}_2 + 2 \text{C} \rightarrow \text{Sn} + 2 \text{CO}
\]
- Classification: Single-replacement (C replaces O in \(\text{SnO}_2\)).
- Final 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 and 1 Pb on the right.
- N: 2 N atoms on the left (in \(\text{Pb(NO}_3\text{)}_2\)) and 2 N atoms on the right (in 2 \(\text{HNO}_3\)).
- O: 6 O atoms on the left (in \(\text{Pb(NO}_3\text{)}_2\)) and 6 O atoms on the right (in 2 \(\text{HNO}_3\)).
- H: 2 H atoms on the left and 2 H atoms on the right (in 2 \(\text{HNO}_3\)).
- S: 1 S atom on the left and 1 S atom on the right (in \(\text{PbS}\)).
\[
\text{Pb(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2 \text{HNO}_3
\]
- Classification: Double-replacement (ions exchange places).
- Final 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 and 1 Hg on the right.
- O: 1 O atom on the left and 2 O atoms on the right (in \(\text{O}_2\)). Multiply \(\text{HgO}\) by 2.
\[
2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2
\]
- Classification: Decomposition (a single compound breaks down into elements).
- Final 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 and 1 K on the right.
- Cl: 1 Cl on the left and 1 Cl on the right.
- O: 3 O atoms on the left and 2 O atoms on the right (in \(\text{O}_2\)). Multiply \(\text{O}_2\) by 3/2 and adjust coefficients:
\[
2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2
\]
- Classification: Decomposition (a single compound breaks down into simpler substances).
- Final 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 atoms on the left and 2 N atoms on the right (in 2 \(\text{NH}_3\)).
- H: 2 H atoms on the left and 6 H atoms on the right (in 2 \(\text{NH}_3\)). Multiply \(\text{H}_2\) by 3.
\[
\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3
\]
- Classification: Synthesis (two elements combine to form a compound).
- Final 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 and 1 Na on the right.
- Br: 1 Br on the left and 2 Br atoms on the right (in \(\text{Br}_2\)). Multiply \(\text{NaBr}\) by 2.
- Cl: 2 Cl atoms on the left (in \(\text{Cl}_2\)) and 2 Cl atoms on the right (in 2 \(\text{NaCl}\)).
\[
2 \text{NaBr} + \text{Cl}_2 \rightarrow 2 \text{NaCl} + \text{Br}_2
\]
- Classification: Single-replacement (Cl replaces Br in \(\text{NaBr}\)).
- Final 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 and 1 Zn on the right.
- Ag: 1 Ag on the right and 2 Ag atoms on the left (in 2 \(\text{AgNO}_3\)). Multiply \(\text{AgNO}_3\) by 2.
- N: 2 N atoms on the right (in \(\text{Zn(NO}_3\text{)}_2\)) and 2 N atoms on the left (in 2 \(\text{AgNO}_3\)).
- O: 6 O atoms on the right (in \(\text{Zn(NO}_3\text{)}_2\)) and 6 O atoms on the left (in 2 \(\text{AgNO}_3\)).
\[
\text{Zn} + 2 \text{AgNO}_3 \rightarrow \text{Zn(NO}_3\text{)}_2 + 2 \text{Ag}
\]
- Classification: Single-replacement (Zn replaces Ag in \(\text{AgNO}_3\)).
- Final 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 and 1 Sn on the right.
- Cl: 2 Cl atoms on the left and 4 Cl atoms on the right (in \(\text{SnCl}_4\)). Multiply \(\text{Cl}_2\) by 2.
\[
\text{Sn} + 2 \text{Cl}_2 \rightarrow \text{SnCl}_4
\]
- Classification: Synthesis (two elements combine to form a compound).
- Final 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 and 1 Ba on the right.
- O: 2 O atoms on the left (in \(\text{Ba(OH)}_2\)) and 2 O atoms on the right (in \(\text{BaO}\) and \(\text{H}_2\text{O}\)).
- H: 2 H atoms on the left and 2 H atoms on the right (in \(\text{H}_2\text{O}\)).
- The equation is already balanced.
- Classification: Decomposition (a single compound breaks down into simpler substances).
- Final Answer: \( \text{Ba(OH)}_2 \rightarrow \text{BaO} + \text{H}_2\text{O} \) (decomposition)

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Final Answer Boxed

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