learning intentions
- i can balance complex chemical equations with multiple steps
- i can classify reaction types and justify my reasoning.
- i can analyze chemical reactions using the law of conservation of mass.
- i can evaluate and correct errors in chemical equations.
reaction types reference
you may use this reference, but you must justify your choice.
- synthesis: a + b → ab
- decomposition: ab → a + b
- single replacement: a + bc → ac + b
- double replacement: ab + cd → ad + cb
- combustion: hydrocarbon + o₂ → co₂ + h₂o
part a: identify, balance, and justify (tuesday)
directions:
- balance each equation
- identify the reaction type.
- explain how you know.
1. ___ al + ___ br₂ → ___ albr₃
balanced equation: ____________________________
reaction type: ___________________
justification: ____________________________
2. ___ fe₂o₃ + ___ h₂ → ___ fe + ___ h₂o
balanced equation: ____________________________
reaction type: ___________________
justification: ____________________________
3. ___ ca + ___ h₂o → ___ ca(oh)₂ + ___ h₂
balanced equation: ____________________________
reaction type: ___________________
justification: ____________________________
part b: multi - step & complex balancing (wednesday)
directions: balance carefully—some elements appear more than once.
4. ___ c₂h₆ + ___ o₂ → ___ co₂ + ___ h₂o
5. ___ kclo₃ → ___ kcl + ___ o₂
6. ___ na₃po₄ + ___ mgcl₂ → ___ mg₃(po₄)₂ + ___ nacl

Question

Accepted Answer

### Problem 1: $\boldsymbol{\ce{Al + Br_{2}
ightarrow AlBr_{3}}}$
# Explanation:
## Step 1: Balance Br atoms
Count Br on each side. Left: 2 (in $\ce{Br_{2}}$), Right: 3 (in $\ce{AlBr_{3}}$). Find LCM of 2 and 3, which is 6. So put 3 in front of $\ce{Br_{2}}$ (3×2=6 Br) and 2 in front of $\ce{AlBr_{3}}$ (2×3=6 Br).
$\ce{Al + 3Br_{2}
ightarrow 2AlBr_{3}}$

## Step 2: Balance Al atoms
Now Al: Left: 1, Right: 2. Put 2 in front of $\ce{Al}$.
$\ce{2Al + 3Br_{2}
ightarrow 2AlBr_{3}}$

## Step 3: Identify reaction type
Synthesis: Two substances (Al, $\ce{Br_{2}}$) combine to form one ($\ce{AlBr_{3}}$), matching $A + B
ightarrow AB$.

# Answer:
Balanced Equation: $\boldsymbol{\ce{2Al + 3Br_{2}
ightarrow 2AlBr_{3}}}$
Reaction Type: Synthesis
Justification: Two reactants (Al and $\ce{Br_{2}}$) combine to form a single product ($\ce{AlBr_{3}}$), fitting the synthesis pattern ($A + B
ightarrow AB$).

### Problem 2: $\boldsymbol{\ce{Fe_{2}O_{3} + H_{2}
ightarrow Fe + H_{2}O}}$
# Explanation:
## Step 1: Balance Fe atoms
Left: 2 Fe (in $\ce{Fe_{2}O_{3}}$), Right: 1 Fe. Put 2 in front of $\ce{Fe}$.
$\ce{Fe_{2}O_{3} + H_{2}
ightarrow 2Fe + H_{2}O}$

## Step 2: Balance O atoms
Left: 3 O (in $\ce{Fe_{2}O_{3}}$), Right: 1 O (in $\ce{H_{2}O}$). Put 3 in front of $\ce{H_{2}O}$.
$\ce{Fe_{2}O_{3} + H_{2}
ightarrow 2Fe + 3H_{2}O}$

## Step 3: Balance H atoms
Right: 6 H (3×2 in $\ce{H_{2}O}$), Left: 2 H (in $\ce{H_{2}}$). Put 3 in front of $\ce{H_{2}}$.
$\ce{Fe_{2}O_{3} + 3H_{2}
ightarrow 2Fe + 3H_{2}O}$

## Step 4: Identify reaction type
Single Replacement: $\ce{H_{2}}$ (A) replaces Fe in $\ce{Fe_{2}O_{3}}$ (BC) to form Fe (B) and $\ce{H_{2}O}$ (AC), matching $A + BC
ightarrow AC + B$.

# Answer:
Balanced Equation: $\boldsymbol{\ce{Fe_{2}O_{3} + 3H_{2}
ightarrow 2Fe + 3H_{2}O}}$
Reaction Type: Single Replacement
Justification: $\ce{H_{2}}$ (a more reactive metal - like element) replaces Fe in $\ce{Fe_{2}O_{3}}$, producing Fe and $\ce{H_{2}O}$, fitting $A + BC
ightarrow AC + B$.

### Problem 3: $\boldsymbol{\ce{Ca + H_{2}O
ightarrow Ca(OH)_{2} + H_{2}}}$
# Explanation:
## Step 1: Balance O atoms
Left: 1 O (in $\ce{H_{2}O}$), Right: 2 O (in $\ce{Ca(OH)_{2}}$). Put 2 in front of $\ce{H_{2}O}$.
$\ce{Ca + 2H_{2}O
ightarrow Ca(OH)_{2} + H_{2}}$

## Step 2: Balance H atoms
Left: 4 H (2×2 in $\ce{H_{2}O}$), Right: 2 (in $\ce{Ca(OH)_{2}}$) + 2 (in $\ce{H_{2}}$) = 4 H. Balanced.

## Step 3: Balance Ca atoms
Left: 1, Right: 1. Already balanced.
$\ce{Ca + 2H_{2}O
ightarrow Ca(OH)_{2} + H_{2}}$

## Step 4: Identify reaction type
Single Replacement: Ca (A) replaces H in $\ce{H_{2}O}$ (BC) to form $\ce{Ca(OH)_{2}}$ (AC) and $\ce{H_{2}}$ (B), matching $A + BC
ightarrow AC + B$.

# Answer:
Balanced Equation: $\boldsymbol{\ce{Ca + 2H_{2}O
ightarrow Ca(OH)_{2} + H_{2}}}$
Reaction Type: Single Replacement
Justification: Ca replaces H in $\ce{H_{2}O}$, forming $\ce{Ca(OH)_{2}}$ and $\ce{H_{2}}$, fitting the single replacement pattern ($A + BC
ightarrow AC + B$).

### Problem 4: $\boldsymbol{\ce{C_{2}H_{6} + O_{2}
ightarrow CO_{2} + H_{2}O}}$ (Assuming $\ce{C_{2}H_{6}}$ is intended)
# Explanation:
## Step 1: Balance C atoms
Left: 2 C (in $\ce{C_{2}H_{6}}$), Right: 1 C (in $\ce{CO_{2}}$). Put 2 in front of $\ce{CO_{2}}$.
$\ce{C_{2}H_{6} + O_{2}
ightarrow 2CO_{2} + H_{2}O}$

## Step 2: Balance H atoms
Left: 6 H (in $\ce{C_{2}H_{6}}$), Right: 2 H (in $\ce{H_{2}O}$). Put 3 in front of $\ce{H_{2}O}$ (3×2=6 H).
$\ce{C_{2}H_{6} + O_{2}
ightarrow 2CO_{2} + 3H_{2}O}$

## Step 3: Balance O atoms
Right: 2×2 (from $\ce{CO_{2}}$) + 3×1 (from $\ce{H_{2}O}$) = 7 O. O is diatomic ($\ce{O_{2}}$), so we need $\frac{7}{2}$ $\ce{O_{2}}$, but multiply all coefficients by 2 to eliminate fractions.
$2\ce{C_{2}H_{6}} + 7\ce{O_{2}}
ightarrow 4\ce{CO_{2}} + 6\ce{H_{2}O}$

## Step 4: Identify reaction type
Combustion: Hydrocarbon ($\ce{C_{2}H_{6}}$) reacts with $\ce{O_{2}}$ to produce $\ce{CO_{2}}$ and $\ce{H_{2}O}$, matching Hydrocarbon + $\ce{O_{2}}
ightarrow \ce{CO_{2}} + \ce{H_{2}O}$.

# Answer:
Balanced Equation: $\boldsymbol{\ce{2C_{2}H_{6} + 7O_{2}
ightarrow 4CO_{2} + 6H_{2}O}}$
Reaction Type: Combustion
Justification: A hydrocarbon ($\ce{C_{2}H_{6}}$) reacts with $\ce{O_{2}}$ to form $\ce{CO_{2}}$ and $\ce{H_{2}O}$, fitting the combustion reaction pattern.

### Problem 5: $\boldsymbol{\ce{KClO_{3}
ightarrow KCl + O_{2}}}$
# Explanation:
## Step 1: Balance O atoms
Left: 3 O (in $\ce{KClO_{3}}$), Right: 2 O (in $\ce{O_{2}}$). LCM of 3 and 2 is 6. Put 2 in front of $\ce{KClO_{3}}$ (2×3=6 O) and 3 in front of $\ce{O_{2}}$ (3×2=6 O).
$\ce{2KClO_{3}
ightarrow KCl + 3O_{2}}$

## Step 2: Balance K and Cl atoms
Left: 2 K, 2 Cl (in $\ce{2KClO_{3}}$), Right: 1 K, 1 Cl (in $\ce{KCl}$). Put 2 in front of $\ce{KCl}$.
$\ce{2KClO_{3}
ightarrow 2KCl + 3O_{2}}$

## Step 3: Identify reaction type
Decomposition: One compound ($\ce{KClO_{3}}$) breaks down into two ($\ce{KCl}$, $\ce{O_{2}}$), matching $AB
ightarrow A + B$.

# Answer:
Balanced Equation: $\boldsymbol{\ce{2KClO_{3}
ightarrow 2KCl + 3O_{2}}}$
Reaction Type: Decomposition
Justification: A single compound ($\ce{KClO_{3}}$) decomposes into two simpler substances ($\ce{KCl}$ and $\ce{O_{2}}$), fitting the decomposition pattern ($AB
ightarrow A + B$).

### Problem 6: $\boldsymbol{\ce{Na_{3}PO_{4} + MgCl_{2}
ightarrow Mg_{3}(PO_{4})_{2} + NaCl}}$ (Assuming $\ce{Na_{3}PO_{4}}$ is intended)
# Explanation:
## Step 1: Balance $\ce{PO_{4}^{3-}}$ ions
Left: 1 $\ce{PO_{4}^{3-}}$ (in $\ce{Na_{3}PO_{4}}$), Right: 2 (in $\ce{Mg_{3}(PO_{4})_{2}}$). Put 2 in front of $\ce{Na_{3}PO_{4}}$.
$\ce{2Na_{3}PO_{4} + MgCl_{2}
ightarrow Mg_{3}(PO_{4})_{2} + NaCl}$

## Step 2: Balance Mg atoms
Left: 1 Mg (in $\ce{MgCl_{2}}$), Right: 3 Mg (in $\ce{Mg_{3}(PO_{4})_{2}}$). Put 3 in front of $\ce{MgCl_{2}}$.
$\ce{2Na_{3}PO_{4} + 3MgCl_{2}
ightarrow Mg_{3}(PO_{4})_{2} + NaCl}$

## Step 3: Balance Na and Cl atoms
Left: 2×3=6 Na, 3×2=6 Cl. Right: 1 Na, 1 Cl (in $\ce{NaCl}$). Put 6 in front of $\ce{NaCl}$.
$\ce{2Na_{3}PO_{4} + 3MgCl_{2}
ightarrow Mg_{3}(PO_{4})_{2} + 6NaCl}$

## Step 4: Identify reaction type
Double Replacement: Cations (Na⁺, Mg²⁺) and anions ($\ce{PO_{4}^{3-}}$, Cl⁻) swap, matching $AB + CD
ightarrow AD + CB$.

# Answer:
Balanced Equation: $\boldsymbol{\ce{2Na_{3}PO_{4} + 3MgCl_{2}
ightarrow Mg_{3}(PO_{4})_{2} + 6NaCl}}$
Reaction Type: Double Replacement
Justification: Ions (Na⁺, Mg²⁺; $\ce{PO_{4}^{3-}}$, Cl⁻) exchange between compounds, fitting $AB + CD
ightarrow AD + CB$.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1: $\boldsymbol{\ce{Al + Br_{2}

Step 1: Balance Br atoms

Step 2: Balance Al atoms

Step 3: Identify reaction type

Step 1: Balance Fe atoms

Step 2: Balance O atoms

Step 3: Balance H atoms

Step 4: Identify reaction type

Step 1: Balance O atoms

Step 2: Balance H atoms

Step 3: Balance Ca atoms

Step 4: Identify reaction type

Answer:

Problem 2: $\boldsymbol{\ce{Fe_{2}O_{3} + H_{2}