Classify And Balance Chemical Equations Exercise Set With Keys ... - Free Printable
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Step-by-step solution for: Classify And Balance Chemical Equations Exercise Set With Keys ...
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Step-by-step solution for: Classify And Balance Chemical Equations Exercise Set With Keys ...
To solve the problem, we need to classify each chemical reaction into its type and balance the equations accordingly. Let's go through each reaction step by step.
---
#### Type of Reaction:
This is a double displacement reaction because the cations and anions switch places.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 + \text{AlCl}_3
\]
- Balance sulfate (\(\text{SO}_4^{2-}\)) first:
- There are 3 \(\text{SO}_4^{2-}\) on the left and 1 on the right. Add a coefficient of 3 to \(\text{BaSO}_4\):
\[
\text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + \text{AlCl}_3
\]
- Balance barium (\(\text{Ba}^{2+}\)):
- There are 3 \(\text{Ba}^{2+}\) on the right and 1 on the left. Add a coefficient of 3 to \(\text{BaCl}_2\):
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + \text{AlCl}_3
\]
- Balance aluminum (\(\text{Al}^{3+}\)):
- There are 2 \(\text{Al}^{3+}\) on the left and 1 on the right. Add a coefficient of 2 to \(\text{AlCl}_3\):
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3
\]
#### Balanced Equation:
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3
\]
---
#### Type of Reaction:
This is a decomposition reaction because a single compound breaks down into simpler substances.
#### Balancing:
- The equation is already balanced as written:
\[
\text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S}
\]
#### Balanced Equation:
\[
\text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S}
\]
---
#### Type of Reaction:
This is a double displacement reaction because the cations and anions switch places.
#### Balancing:
- Correct the products: The product should be \(\text{Na}_2\text{SO}_4\) instead of \(\text{Na}_2\text{PO}_4\).
- Start with the corrected unbalanced equation:
\[
\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
- Balance sodium (\(\text{Na}^+\)):
- There are 2 \(\text{Na}^+\) on the right and 1 on the left. Add a coefficient of 2 to \(\text{NaOH}\):
\[
2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
- Check all atoms:
- Sodium (\(\text{Na}\)): 2 on both sides.
- Oxygen (\(\text{O}\)): 6 on both sides.
- Hydrogen (\(\text{H}\)): 2 on both sides.
- Copper (\(\text{Cu}\)): 1 on both sides.
- Sulfur (\(\text{S}\)): 1 on both sides.
#### Balanced Equation:
\[
2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
---
#### Type of Reaction:
This is a single displacement reaction because a metal (Fe) displaces hydrogen from an acid.
#### Balancing:
- Correct the product: The product should be \(\text{FeSO}_4\) instead of \(\text{Fe}_2\text{SO}_4\).
- Start with the corrected unbalanced equation:
\[
\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2
\]
- Balance iron (\(\text{Fe}\)):
- There is 1 \(\text{Fe}\) on the left and 1 on the right. No change needed.
- Balance hydrogen (\(\text{H}\)):
- There are 2 \(\text{H}\) on the right and 2 on the left. No change needed.
- Balance sulfur (\(\text{S}\)) and oxygen (\(\text{O}\)):
- Both are already balanced.
#### Balanced Equation:
\[
\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2
\]
---
#### Type of Reaction:
This is a combustion reaction because a hydrocarbon reacts with oxygen to produce carbon dioxide and water.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- Balance carbon (\(\text{C}\)):
- There are 4 \(\text{C}\) on the left and 1 on the right. Add a coefficient of 4 to \(\text{CO}_2\):
\[
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + \text{H}_2\text{O}
\]
- Balance hydrogen (\(\text{H}\)):
- There are 10 \(\text{H}\) on the left and 2 on the right. Add a coefficient of 5 to \(\text{H}_2\text{O}\):
\[
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}
\]
- Balance oxygen (\(\text{O}\)):
- There are 13 \(\text{O}\) on the right (8 in \(\text{CO}_2\) and 5 in \(\text{H}_2\text{O}\)) and 2 on the left. Add a coefficient of \(\frac{13}{2}\) to \(\text{O}_2\). To avoid fractions, multiply everything by 2:
\[
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
\]
#### Balanced Equation:
\[
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
\]
---
#### Type of Reaction:
This is a combustion reaction because a compound reacts with oxygen to produce other compounds.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{H}_2\text{S} + \text{O}_2 \rightarrow \text{SO}_2 + \text{H}_2\text{O}
\]
- Balance sulfur (\(\text{S}\)):
- There is 1 \(\text{S}\) on the left and 1 on the right. No change needed.
- Balance hydrogen (\(\text{H}\)):
- There are 2 \(\text{H}\) on the left and 2 on the right. No change needed.
- Balance oxygen (\(\text{O}\)):
- There are 2 \(\text{O}\) on the left and 3 on the right. Add a coefficient of 2 to \(\text{O}_2\) and adjust \(\text{H}_2\text{O}\):
\[
2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O}
\]
#### Balanced Equation:
\[
2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O}
\]
---
#### Type of Reaction:
This is a combustion reaction because a compound reacts with oxygen to produce carbon dioxide and water.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- Balance carbon (\(\text{C}\)):
- There are 5 \(\text{C}\) on the left and 1 on the right. Add a coefficient of 5 to \(\text{CO}_2\):
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow 5\text{CO}_2 + \text{H}_2\text{O}
\]
- Balance hydrogen (\(\text{H}\)):
- There are 10 \(\text{H}\) on the left and 2 on the right. Add a coefficient of 5 to \(\text{H}_2\text{O}\):
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
- Balance oxygen (\(\text{O}\)):
- There are 6 \(\text{O}\) on the left (5 in \(\text{CO}_2\) and 5 in \(\text{H}_2\text{O}\)) and 1 on the left. Add a coefficient of 6 to \(\text{O}_2\):
\[
\text{C}_5\text{H}_{10}\text{O} + 6\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
#### Balanced Equation:
\[
\text{C}_5\text{H}_{10}\text{O} + 6\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
---
#### Type of Reaction:
This is a single displacement reaction because a more reactive metal (Al) displaces a less reactive metal (Ni).
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Al} + \text{NiBr}_2 \rightarrow \text{AlBr}_3 + \text{Ni}
\]
- Balance aluminum (\(\text{Al}\)):
- There is 1 \(\text{Al}\) on the left and 1 on the right. No change needed.
- Balance bromine (\(\text{Br}\)):
- There are 2 \(\text{Br}\) on the left and 3 on the right. Add a coefficient of 2 to \(\text{AlBr}_3\) and 3 to \(\text{NiBr}_2\):
\[
2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni}
\]
#### Balanced Equation:
\[
2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni}
\]
---
#### Type of Reaction:
This is a synthesis reaction because two elements combine to form a compound.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3
\]
- Balance aluminum (\(\text{Al}\)):
- There are 2 \(\text{Al}\) on the right and 1 on the left. Add a coefficient of 2 to \(\text{Al}\):
\[
2\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3
\]
- Balance oxygen (\(\text{O}\)):
- There are 3 \(\text{O}\) on the right and 2 on the left. Add a coefficient of \(\frac{3}{2}\) to \(\text{O}_2\). To avoid fractions, multiply everything by 2:
\[
4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
\]
#### Balanced Equation:
\[
4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
\]
---
#### Type of Reaction:
This is a decomposition reaction because a single compound breaks down into simpler substances.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2
\]
- Balance hydrogen (\(\text{H}\)):
- There are 2 \(\text{H}\) on the left and 2 on the right. No change needed.
- Balance oxygen (\(\text{O}\)):
- There are 2 \(\text{O}\) on the left and 3 on the right. Add a coefficient of 2 to \(\text{H}_2\text{O}\):
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2
\]
#### Balanced Equation:
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2
\]
---
#### Type of Reaction:
This is a synthesis reaction because two elements combine to form a compound.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{K} + \text{Cl}_2 \rightarrow \text{KCl}
\]
- Balance potassium (\(\text{K}\)):
- There is 1 \(\text{K}\) on the left and 1 on the right. No change needed.
- Balance chlorine (\(\text{Cl}\)):
- There are 2 \(\text{Cl}\) on the left and 1 on the right. Add a coefficient of 2 to \(\text{KCl}\):
\[
2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl}
\]
#### Balanced Equation:
\[
2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl}
\]
---
#### Type of Reaction:
This is a single displacement reaction because a more reactive metal (Na) displaces a less reactive metal (Mg).
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Na} + \text{MgCl}_2 \rightarrow \text{NaCl} + \text{Mg}
\]
- Balance sodium (\(\text{Na}\)):
- There is 1 \(\text{Na}\) on the left and 1 on the right. No change needed.
- Balance magnesium (\(\text{Mg}\)):
- There is 1 \(\text{Mg}\) on the left and 1 on the right. No change needed.
- Balance chlorine (\(\text{Cl}\)):
- There are 2 \(\text{Cl}\) on the left and 1 on the right. Add a coefficient of 2 to \(\text{NaCl}\):
\[
2\text{Na} + \text{MgCl}_2 \rightarrow 2\text{NaCl} + \text{Mg}
\]
#### Balanced Equation:
\[
2\text{Na} + \text{MgCl}_2 \rightarrow 2\text{NaCl} + \text{Mg}
\]
---
\[
\boxed{
\begin{array}{ll}
(1) & \text{Double displacement; } \text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3 \\
(2) & \text{Decomposition; } \text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S} \\
(3) & \text{Double displacement; } 2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2 \\
(4) & \text{Single displacement; } \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2 \\
(5) & \text{Combustion; } 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \\
(6) & \text{Combustion; } 2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O} \\
(7) & \text{Combustion; } \text{C}_5\text{H}_{10}\text{O} + 6\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O} \\
(8) & \text{Single displacement; } 2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni} \\
(9) & \text{Synthesis; } 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \\
(10) & \text{Decomposition; } 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \\
(11) & \text{Synthesis; } 2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl} \\
(12) & \text{Single displacement; } 2\text{Na} + \text{MgCl}_2 \rightarrow 2\text{NaCl} + \text{Mg} \\
\end{array}
}
\]
---
(1) \( \text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 + \text{AlCl}_3 \)
#### Type of Reaction:
This is a double displacement reaction because the cations and anions switch places.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 + \text{AlCl}_3
\]
- Balance sulfate (\(\text{SO}_4^{2-}\)) first:
- There are 3 \(\text{SO}_4^{2-}\) on the left and 1 on the right. Add a coefficient of 3 to \(\text{BaSO}_4\):
\[
\text{Al}_2(\text{SO}_4)_3 + \text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + \text{AlCl}_3
\]
- Balance barium (\(\text{Ba}^{2+}\)):
- There are 3 \(\text{Ba}^{2+}\) on the right and 1 on the left. Add a coefficient of 3 to \(\text{BaCl}_2\):
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + \text{AlCl}_3
\]
- Balance aluminum (\(\text{Al}^{3+}\)):
- There are 2 \(\text{Al}^{3+}\) on the left and 1 on the right. Add a coefficient of 2 to \(\text{AlCl}_3\):
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3
\]
#### Balanced Equation:
\[
\text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3
\]
---
(2) \( \text{Al}_2\text{S}_3 \rightarrow \text{Al} + \text{S} \)
#### Type of Reaction:
This is a decomposition reaction because a single compound breaks down into simpler substances.
#### Balancing:
- The equation is already balanced as written:
\[
\text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S}
\]
#### Balanced Equation:
\[
\text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S}
\]
---
(3) \( \text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{PO}_4 + \text{Cu(OH)}_2 \)
#### Type of Reaction:
This is a double displacement reaction because the cations and anions switch places.
#### Balancing:
- Correct the products: The product should be \(\text{Na}_2\text{SO}_4\) instead of \(\text{Na}_2\text{PO}_4\).
- Start with the corrected unbalanced equation:
\[
\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
- Balance sodium (\(\text{Na}^+\)):
- There are 2 \(\text{Na}^+\) on the right and 1 on the left. Add a coefficient of 2 to \(\text{NaOH}\):
\[
2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
- Check all atoms:
- Sodium (\(\text{Na}\)): 2 on both sides.
- Oxygen (\(\text{O}\)): 6 on both sides.
- Hydrogen (\(\text{H}\)): 2 on both sides.
- Copper (\(\text{Cu}\)): 1 on both sides.
- Sulfur (\(\text{S}\)): 1 on both sides.
#### Balanced Equation:
\[
2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2
\]
---
(4) \( \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2\text{SO}_4 + \text{H}_2 \)
#### Type of Reaction:
This is a single displacement reaction because a metal (Fe) displaces hydrogen from an acid.
#### Balancing:
- Correct the product: The product should be \(\text{FeSO}_4\) instead of \(\text{Fe}_2\text{SO}_4\).
- Start with the corrected unbalanced equation:
\[
\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2
\]
- Balance iron (\(\text{Fe}\)):
- There is 1 \(\text{Fe}\) on the left and 1 on the right. No change needed.
- Balance hydrogen (\(\text{H}\)):
- There are 2 \(\text{H}\) on the right and 2 on the left. No change needed.
- Balance sulfur (\(\text{S}\)) and oxygen (\(\text{O}\)):
- Both are already balanced.
#### Balanced Equation:
\[
\text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2
\]
---
(5) \( \text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
#### Type of Reaction:
This is a combustion reaction because a hydrocarbon reacts with oxygen to produce carbon dioxide and water.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- Balance carbon (\(\text{C}\)):
- There are 4 \(\text{C}\) on the left and 1 on the right. Add a coefficient of 4 to \(\text{CO}_2\):
\[
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + \text{H}_2\text{O}
\]
- Balance hydrogen (\(\text{H}\)):
- There are 10 \(\text{H}\) on the left and 2 on the right. Add a coefficient of 5 to \(\text{H}_2\text{O}\):
\[
\text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}
\]
- Balance oxygen (\(\text{O}\)):
- There are 13 \(\text{O}\) on the right (8 in \(\text{CO}_2\) and 5 in \(\text{H}_2\text{O}\)) and 2 on the left. Add a coefficient of \(\frac{13}{2}\) to \(\text{O}_2\). To avoid fractions, multiply everything by 2:
\[
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
\]
#### Balanced Equation:
\[
2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}
\]
---
(6) \( \text{H}_2\text{S} + \text{O}_2 \rightarrow \text{SO}_2 + \text{H}_2\text{O} \)
#### Type of Reaction:
This is a combustion reaction because a compound reacts with oxygen to produce other compounds.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{H}_2\text{S} + \text{O}_2 \rightarrow \text{SO}_2 + \text{H}_2\text{O}
\]
- Balance sulfur (\(\text{S}\)):
- There is 1 \(\text{S}\) on the left and 1 on the right. No change needed.
- Balance hydrogen (\(\text{H}\)):
- There are 2 \(\text{H}\) on the left and 2 on the right. No change needed.
- Balance oxygen (\(\text{O}\)):
- There are 2 \(\text{O}\) on the left and 3 on the right. Add a coefficient of 2 to \(\text{O}_2\) and adjust \(\text{H}_2\text{O}\):
\[
2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O}
\]
#### Balanced Equation:
\[
2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O}
\]
---
(7) \( \text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \)
#### Type of Reaction:
This is a combustion reaction because a compound reacts with oxygen to produce carbon dioxide and water.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}
\]
- Balance carbon (\(\text{C}\)):
- There are 5 \(\text{C}\) on the left and 1 on the right. Add a coefficient of 5 to \(\text{CO}_2\):
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow 5\text{CO}_2 + \text{H}_2\text{O}
\]
- Balance hydrogen (\(\text{H}\)):
- There are 10 \(\text{H}\) on the left and 2 on the right. Add a coefficient of 5 to \(\text{H}_2\text{O}\):
\[
\text{C}_5\text{H}_{10}\text{O} + \text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
- Balance oxygen (\(\text{O}\)):
- There are 6 \(\text{O}\) on the left (5 in \(\text{CO}_2\) and 5 in \(\text{H}_2\text{O}\)) and 1 on the left. Add a coefficient of 6 to \(\text{O}_2\):
\[
\text{C}_5\text{H}_{10}\text{O} + 6\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
#### Balanced Equation:
\[
\text{C}_5\text{H}_{10}\text{O} + 6\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O}
\]
---
(8) \( \text{Al} + \text{NiBr}_2 \rightarrow \text{AlBr}_3 + \text{Ni} \)
#### Type of Reaction:
This is a single displacement reaction because a more reactive metal (Al) displaces a less reactive metal (Ni).
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Al} + \text{NiBr}_2 \rightarrow \text{AlBr}_3 + \text{Ni}
\]
- Balance aluminum (\(\text{Al}\)):
- There is 1 \(\text{Al}\) on the left and 1 on the right. No change needed.
- Balance bromine (\(\text{Br}\)):
- There are 2 \(\text{Br}\) on the left and 3 on the right. Add a coefficient of 2 to \(\text{AlBr}_3\) and 3 to \(\text{NiBr}_2\):
\[
2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni}
\]
#### Balanced Equation:
\[
2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni}
\]
---
(9) \( \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \)
#### Type of Reaction:
This is a synthesis reaction because two elements combine to form a compound.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3
\]
- Balance aluminum (\(\text{Al}\)):
- There are 2 \(\text{Al}\) on the right and 1 on the left. Add a coefficient of 2 to \(\text{Al}\):
\[
2\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3
\]
- Balance oxygen (\(\text{O}\)):
- There are 3 \(\text{O}\) on the right and 2 on the left. Add a coefficient of \(\frac{3}{2}\) to \(\text{O}_2\). To avoid fractions, multiply everything by 2:
\[
4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
\]
#### Balanced Equation:
\[
4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3
\]
---
(10) \( \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2 \)
#### Type of Reaction:
This is a decomposition reaction because a single compound breaks down into simpler substances.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \text{O}_2
\]
- Balance hydrogen (\(\text{H}\)):
- There are 2 \(\text{H}\) on the left and 2 on the right. No change needed.
- Balance oxygen (\(\text{O}\)):
- There are 2 \(\text{O}\) on the left and 3 on the right. Add a coefficient of 2 to \(\text{H}_2\text{O}\):
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2
\]
#### Balanced Equation:
\[
2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2
\]
---
(11) \( \text{K} + \text{Cl}_2 \rightarrow \text{KCl} \)
#### Type of Reaction:
This is a synthesis reaction because two elements combine to form a compound.
#### Balancing:
- Start with the unbalanced equation:
\[
\text{K} + \text{Cl}_2 \rightarrow \text{KCl}
\]
- Balance potassium (\(\text{K}\)):
- There is 1 \(\text{K}\) on the left and 1 on the right. No change needed.
- Balance chlorine (\(\text{Cl}\)):
- There are 2 \(\text{Cl}\) on the left and 1 on the right. Add a coefficient of 2 to \(\text{KCl}\):
\[
2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl}
\]
#### Balanced Equation:
\[
2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl}
\]
---
(12) \( \text{Na} + \text{MgCl}_2 \rightarrow \text{NaCl} + \text{Mg} \)
#### Type of Reaction:
This is a single displacement reaction because a more reactive metal (Na) displaces a less reactive metal (Mg).
#### Balancing:
- Start with the unbalanced equation:
\[
\text{Na} + \text{MgCl}_2 \rightarrow \text{NaCl} + \text{Mg}
\]
- Balance sodium (\(\text{Na}\)):
- There is 1 \(\text{Na}\) on the left and 1 on the right. No change needed.
- Balance magnesium (\(\text{Mg}\)):
- There is 1 \(\text{Mg}\) on the left and 1 on the right. No change needed.
- Balance chlorine (\(\text{Cl}\)):
- There are 2 \(\text{Cl}\) on the left and 1 on the right. Add a coefficient of 2 to \(\text{NaCl}\):
\[
2\text{Na} + \text{MgCl}_2 \rightarrow 2\text{NaCl} + \text{Mg}
\]
#### Balanced Equation:
\[
2\text{Na} + \text{MgCl}_2 \rightarrow 2\text{NaCl} + \text{Mg}
\]
---
Final Answer:
\[
\boxed{
\begin{array}{ll}
(1) & \text{Double displacement; } \text{Al}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \rightarrow 3\text{BaSO}_4 + 2\text{AlCl}_3 \\
(2) & \text{Decomposition; } \text{Al}_2\text{S}_3 \rightarrow 2\text{Al} + 3\text{S} \\
(3) & \text{Double displacement; } 2\text{NaOH} + \text{CuSO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Cu(OH)}_2 \\
(4) & \text{Single displacement; } \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2 \\
(5) & \text{Combustion; } 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O} \\
(6) & \text{Combustion; } 2\text{H}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{SO}_2 + 2\text{H}_2\text{O} \\
(7) & \text{Combustion; } \text{C}_5\text{H}_{10}\text{O} + 6\text{O}_2 \rightarrow 5\text{CO}_2 + 5\text{H}_2\text{O} \\
(8) & \text{Single displacement; } 2\text{Al} + 3\text{NiBr}_2 \rightarrow 2\text{AlBr}_3 + 3\text{Ni} \\
(9) & \text{Synthesis; } 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \\
(10) & \text{Decomposition; } 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \\
(11) & \text{Synthesis; } 2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl} \\
(12) & \text{Single displacement; } 2\text{Na} + \text{MgCl}_2 \rightarrow 2\text{NaCl} + \text{Mg} \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of balancing chemical reactions worksheet with answers.
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/balancing-chemical-reactions-worksheet-with-answers/balancing-chemical-reactions-worksheet-with-answers-9.jpg)
![49 Balancing Chemical Equations Worksheets [with Answers]](/images2/balancing-chemical-reactions-worksheet-with-answers/balancing-chemical-reactions-worksheet-with-answers-5.jpg)
