Question

balancing and identifying chemical equations
classify the following reactions and balance them.
(1) ( \text{al}_2(\text{so}_4)_3 + \text{bacl}_2
ightarrow \text{baso}_4 + \text{alcl}_3 ) \t type of reaction:
(2) ( \text{al}_2\text{s}_3
ightarrow \text{al} + \text{s} ) \t type of reaction:
(3) ( \text{naoh} + \text{cuso}_4
ightarrow \text{na}_2\text{po}_4 + \text{cu(oh)}_2 ) \t type of reaction:
(4) ( \text{fe} + \text{h}_2\text{so}_4
ightarrow \text{fe}_2\text{so}_4 + \text{h}_2 ) \t type of reaction:
(5) ( \text{c}_4\text{h}_{12} + \text{o}_2
ightarrow \text{co}_2 + \text{h}_2\text{o} ) \t type of reaction:
(6) ( \text{h}_2\text{s} + \text{o}_2
ightarrow \text{so}_2 + \text{h}_2\text{o} ) \t type of reaction:
(7) ( \text{c}_5\text{h}_9\text{o} + \text{o}_2
ightarrow \text{co}_2 + \text{h}_2\text{o} ) \t type of reaction
(8) ( \text{al} + \text{nibr}_2
ightarrow \text{albr}_3 + \text{ni} ) \t type of reaction
( \text{al} + \text{o}_2
ightarrow \text{al}_2\text{o}_3 ) \t type of reaction
( \text{h}_2\text{o}_2
ightarrow \text{h}_2\text{o} + \text{o}_2 ) \t type of reaction
( \text{cl}_2
ightarrow \text{kcl} ) \t type of reaction

Explanation:

Step1: Analyze Reaction (1)

Reaction: $\ce{Al_{2}(SO_{4})_{3} + BaCl_{2} -> BaSO_{4} + AlCl_{3}}$
This is a double - displacement reaction (ions exchange). To balance:
- $\ce{SO_4^{2 - }}$: 3 on left, so $\ce{BaSO_4}$ needs 3.
- $\ce{Ba^{2 + }}$: 3 $\ce{BaSO_4}$ means 3 $\ce{BaCl_2}$.
- $\ce{Cl^-}$: 3 $\ce{BaCl_2}$ gives 6 $\ce{Cl^-}$, so $\ce{AlCl_3}$ needs 2 (6 $\ce{Cl^-}$).
- $\ce{Al^{3 + }}$: 2 $\ce{AlCl_3}$ means 1 $\ce{Al_{2}(SO_{4})_{3}}$.

Balanced: $\ce{Al_{2}(SO_{4})_{3} + 3BaCl_{2} = 3BaSO_{4} + 2AlCl_{3}}$
Type: Double - Displacement.

Step2: Analyze Reaction (2)

Reaction: $\ce{Al_{2}S_{3} -> Al + S}$
This is a decomposition reaction (compound breaks into elements). To balance:
- $\ce{Al}$: 2 on left, so 2 $\ce{Al}$ on right.
- $\ce{S}$: 3 on left, so 3 $\ce{S}$ on right.

Balanced: $\ce{Al_{2}S_{3} = 2Al + 3S}$
Type: Decomposition.

Step3: Analyze Reaction (3)

Reaction: $\ce{NaOH + CuSO_{4} -> Na_{2}SO_{4} + Cu(OH)_{2}}$ (Assuming typo: $\ce{Na_{2}PO_{4}}$ should be $\ce{Na_{2}SO_{4}}$)
Double - displacement (ions exchange). Balance:
- $\ce{Na^+}$: 2 on right, so 2 $\ce{NaOH}$.
- $\ce{OH^-}$: 2 $\ce{NaOH}$ gives 2 $\ce{OH^-}$, matching $\ce{Cu(OH)_{2}}$.
- $\ce{SO_4^{2 - }}$: 1 on left and right.
- $\ce{Cu^{2 + }}$: 1 on left and right.

Balanced: $\ce{2NaOH + CuSO_{4} = Na_{2}SO_{4} + Cu(OH)_{2}\downarrow}$
Type: Double - Displacement.

Step4: Analyze Reaction (4)

Reaction: $\ce{Fe + H_{2}SO_{4} -> FeSO_{4} + H_{2}}$ (Assuming typo: $\ce{Fe_{2}SO_{4}}$ should be $\ce{FeSO_{4}}$)
This is a single - displacement reaction (Fe displaces H). Balance:
- $\ce{Fe}$: 1 on left and right.
- $\ce{H}$: 2 on left and right.
- $\ce{SO_4^{2 - }}$: 1 on left and right.

Balanced: $\ce{Fe + H_{2}SO_{4} = FeSO_{4} + H_{2}\uparrow}$
Type: Single - Displacement.

Step5: Analyze Reaction (5)

Reaction: $\ce{C_{4}H_{12} + O_{2} -> CO_{2} + H_{2}O}$
This is a combustion reaction (hydrocarbon + $\ce{O_2}$). Balance:
- $\ce{C}$: 4 on left, so 4 $\ce{CO_2}$.
- $\ce{H}$: 12 on left, so 6 $\ce{H_2O}$ (12 H).
- $\ce{O}$: Right: $4\times2 + 6\times1=14$, so $\ce{O_2}$: 7 (14 O).

Balanced: $\ce{C_{4}H_{12} + 7O_{2} = 4CO_{2} + 6H_{2}O}$
Type: Combustion.

Step6: Analyze Reaction (6)

Reaction: $\ce{H_{2}S + O_{2} -> SO_{2} + H_{2}O}$
Combustion (or oxidation - reduction). Balance:
- $\ce{H}$: 2 on left, so 1 $\ce{H_2O}$.
- $\ce{S}$: 1 on left, so 1 $\ce{SO_2}$.
- $\ce{O}$: Right: $2 + 1 = 3$, left $\ce{O_2}$: $\frac{3}{2}$, multiply by 2:
$2\ce{H_{2}S + 3O_{2} = 2SO_{2} + 2H_{2}O}$

Type: Combustion (Oxidation - Reduction).

Step7: Analyze Reaction (7)

Reaction: $\ce{C_{5}H_{9}O + O_{2} -> CO_{2} + H_{2}O}$
Combustion. Balance:
- $\ce{C}$: 5 on left, so 5 $\ce{CO_2}$.
- $\ce{H}$: 9 on left, so $\frac{9}{2}\ce{H_2O}$, multiply by 2: 2 $\ce{C_{5}H_{9}O}$, 10 $\ce{CO_2}$, 9 $\ce{H_2O}$.
- $\ce{O}$: Left: $2\times1 + O_2$; Right: $10\times2+9\times1 = 29$. So $\ce{O_2}$: $\frac{29 - 2}{2}= \frac{27}{2}$, multiply by 2:
$2\ce{C_{5}H_{9}O + 27O_{2} = 10CO_{2} + 9H_{2}O}$

Type: Combustion.

Step8: Analyze Reaction (8)

Reaction: $\ce{Al + NiBr_{2} -> AlBr_{3} + Ni}$
Single - displacement (Al displaces Ni). Balance:
- $\ce{Br}$: 3 on right, 2 on left. LCM(2,3)=6. So 3 $\ce{NiBr_{2}}$, 2 $\ce{AlBr_{3}}$.
- $\ce{Ni}$: 3 on left, so 3 $\ce{Ni}$ on right.
- $\ce{Al}$: 2 on right, so 2 $\ce{Al}$ on left.

Balanced: $\ce{2Al + 3NiBr_{2} = 2AlBr_{3} + 3Ni}$
Type: Single - Displacement.

Step9: Analyze Reaction (9)

Reaction: $\ce{Al + O_{2} -> Al_{2}O_{3}}$
Synthesis (elements form compound). Balance:
- $\ce{Al}$: 2 on right, so 2 $\ce{Al}$ on left (first), then 4 $\ce{Al}$ (to balance O).
- $\ce{O}$: 3 on right, 2 on left. LCM(2,3)=6. So 3 $\ce{O_2}$, 2 $\ce{Al_2O_3}$, 4 $\ce{Al}$.

Balanced: $\ce{4Al + 3O_{2} = 2Al_{2}O_{3}}$
Type: Synthesis.

Step10: Analyze Reaction (10)

Reaction: $\ce{H_{2}O_{2} -> H_{2}O + O_{2}}$
Decomposition. Balance:
- $\ce{H}$: 2 on left, so 2 $\ce{H_2O}$.
- $\ce{O}$: Left: 2; Right: $2 + 2 = 4$. So 2 $\ce{H_2O_2}$, 2 $\ce{H_2O}$, 1 $\ce{O_2}$.

Balanced: $\ce{2H_{2}O_{2} = 2H_{2}O + O_{2}\uparrow}$
Type: Decomposition.

Step11: Analyze Reaction (11)

Assuming reaction: $\ce{K + Cl_{2} -> KCl}$ (typo in original). Synthesis. Balance:
- $\ce{Cl}$: 2 on left, so 2 $\ce{KCl}$, 2 $\ce{K}$.

Balanced: $\ce{2K + Cl_{2} = 2KCl}$
Type: Synthesis.

Answer:

(1) Balanced: $\boldsymbol{\ce{Al_{2}(SO_{4})_{3} + 3BaCl_{2} = 3BaSO_{4} + 2AlCl_{3}}}$, Type: Double - Displacement
(2) Balanced: $\boldsymbol{\ce{Al_{2}S_{3} = 2Al + 3S}}$, Type: Decomposition
(3) Balanced: $\boldsymbol{\ce{2NaOH + CuSO_{4} = Na_{2}SO_{4} + Cu(OH)_{2}\downarrow}}$ (corrected), Type: Double - Displacement
(4) Balanced: $\boldsymbol{\ce{Fe + H_{2}SO_{4} = FeSO_{4} + H_{2}\uparrow}}$ (corrected), Type: Single - Displacement
(5) Balanced: $\boldsymbol{\ce{C_{4}H_{12} + 7O_{2} = 4CO_{2} + 6H_{2}O}}$, Type: Combustion
(6) Balanced: $\boldsymbol{\ce{2H_{2}S + 3O_{2} = 2SO_{2} + 2H_{2}O}}$, Type: Combustion
(7) Balanced: $\boldsymbol{\ce{2C_{5}H_{9}O + 27O_{2} = 10CO_{2} + 9H_{2}O}}$, Type: Combustion
(8) Balanced: $\boldsymbol{\ce{2Al + 3NiBr_{2} = 2AlBr_{3} + 3Ni}}$, Type: Single - Displacement
(9) Balanced: $\boldsymbol{\ce{4Al + 3O_{2} = 2Al_{2}O_{3}}}$, Type: Synthesis
(10) Balanced: $\boldsymbol{\ce{2H_{2}O_{2} = 2H_{2}O + O_{2}\uparrow}}$, Type: Decomposition
(11) Balanced: $\boldsymbol{\ce{2K + Cl_{2} = 2KCl}}$ (corrected), Type: Synthesis