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### Problem 1: Balancing $ \boldsymbol{\ce{NaBr + Ca(OH)2 -> CaBr2 + NaOH}} $ and Identifying Reaction Type
# Explanation:
## Step 1: Balance Br atoms
On the left, we have 1 Br (from $ \ce{NaBr} $), and on the right, we have 2 Br (from $ \ce{CaBr2} $). So we put a coefficient of 2 in front of $ \ce{NaBr} $:
$ 2\ce{NaBr} + \ce{Ca(OH)2} -> \ce{CaBr2} + \ce{NaOH} $

## Step 2: Balance Na atoms
Now we have 2 Na on the left (from $ 2\ce{NaBr} $), so we put a coefficient of 2 in front of $ \ce{NaOH} $ to balance Na:
$ 2\ce{NaBr} + \ce{Ca(OH)2} -> \ce{CaBr2} + 2\ce{NaOH} $

## Step 3: Check other atoms (Ca, O, H)
- Ca: 1 on left ($ \ce{Ca(OH)2} $) and 1 on right ($ \ce{CaBr2} $) – balanced.
- O: 2 on left ($ \ce{Ca(OH)2} $) and 2 on right ($ 2\ce{NaOH} $) – balanced.
- H: 2 on left ($ \ce{Ca(OH)2} $) and 2 on right ($ 2\ce{NaOH} $) – balanced.

## Step 4: Identify reaction type
This is a double replacement (DR) reaction because the cations ($ \ce{Na+} $ and $ \ce{Ca^{2+}} $) and anions ($ \ce{Br-} $ and $ \ce{OH-} $) switch partners.

# Answer:
Balanced equation: $ \boldsymbol{2\ce{NaBr} + \ce{Ca(OH)2} -> \ce{CaBr2} + 2\ce{NaOH}} $
Reaction type: $ \boldsymbol{DR} $

### Problem 2: Balancing $ \boldsymbol{\ce{NH3 + H2SO4 -> (NH4)2SO4}} $ and Identifying Reaction Type
# Explanation:
## Step 1: Balance $ \ce{NH4+} $ (from $ \ce{NH3} $ and $ \ce{(NH4)2SO4} $)
On the right, we have 2 $ \ce{NH4+} $ (from $ \ce{(NH4)2SO4} $), so we put a coefficient of 2 in front of $ \ce{NH3} $:
$ 2\ce{NH3} + \ce{H2SO4} -> \ce{(NH4)2SO4} $

## Step 2: Check other atoms (S, O, H)
- S: 1 on left ($ \ce{H2SO4} $) and 1 on right ($ \ce{(NH4)2SO4} $) – balanced.
- O: 4 on left ($ \ce{H2SO4} $) and 4 on right ($ \ce{(NH4)2SO4} $) – balanced.
- H: Let’s verify:
  - Left: $ 2\ce{NH3} $ has $ 2 \times 3 = 6 $ H, $ \ce{H2SO4} $ has 2 H – total 8 H.
  - Right: $ \ce{(NH4)2SO4} $ has $ 2 \times 4 = 8 $ H – balanced.

## Step 3: Identify reaction type
This is a synthesis (S) reaction because two substances ($ \ce{NH3} $ and $ \ce{H2SO4} $) combine to form a single product ($ \ce{(NH4)2SO4} $).

# Answer:
Balanced equation: $ \boldsymbol{2\ce{NH3} + \ce{H2SO4} -> \ce{(NH4)2SO4}} $
Reaction type: $ \boldsymbol{S} $

### Problem 3: Balancing $ \boldsymbol{\ce{Pb + H3PO4 -> H2 + Pb3(PO4)2}} $ and Identifying Reaction Type
# Explanation:
## Step 1: Balance $ \ce{PO4^{3-}} $ (from $ \ce{H3PO4} $ and $ \ce{Pb3(PO4)2} $)
On the right, we have 2 $ \ce{PO4^{3-}} $ (from $ \ce{Pb3(PO4)2} $), so we put a coefficient of 2 in front of $ \ce{H3PO4} $:
$ \ce{Pb} + 2\ce{H3PO4} -> \ce{H2} + \ce{Pb3(PO4)2} $

## Step 2: Balance Pb atoms
On the right, we have 3 Pb (from $ \ce{Pb3(PO4)2} $), so we put a coefficient of 3 in front of $ \ce{Pb} $:
$ 3\ce{Pb} + 2\ce{H3PO4} -> \ce{H2} + \ce{Pb3(PO4)2} $

## Step 3: Balance H atoms
On the left, $ 2\ce{H3PO4} $ has $ 2 \times 3 = 6 $ H. On the right, $ \ce{H2} $ has 2 H per molecule. Let’s find the coefficient for $ \ce{H2} $:
$ 6 $ H on left $ = 2 \times $ (coefficient of $ \ce{H2} $) → coefficient of $ \ce{H2} = 3 $:
$ 3\ce{Pb} + 2\ce{H3PO4} -> 3\ce{H2} + \ce{Pb3(PO4)2} $

## Step 4: Check O atoms
Left: $ 2\ce{H3PO4} $ has $ 2 \times 4 = 8 $ O. Right: $ \ce{Pb3(PO4)2} $ has $ 2 \times 4 = 8 $ O – balanced.

## Step 5: Identify reaction type
This is a single replacement (SR) reaction because $ \ce{Pb} $ (a metal) replaces $ \ce{H} $ in $ \ce{H3PO4} $ (an acid) to form $ \ce{H2} $ gas.

# Answer:
Balanced equation: $ \boldsymbol{3\ce{Pb} + 2\ce{H3PO4} -> 3\ce{H2} + \ce{Pb3(PO4)2}} $
Reaction type: $ \boldsymbol{SR} $

### Problem 4: Balancing $ \boldsymbol{\ce{Li3N + NH4NO3 -> LiNO3 + (NH4)3N}} $ and Identifying Reaction Type
# Explanation:
## Step 1: Balance $ \ce{NH4+} $ (from $ \ce{NH4NO3} $ and $ \ce{(NH4)3N} $)
On the right, $ \ce{(NH4)3N} $ has 3 $ \ce{NH4+} $, so we put a coefficient of 3 in front of $ \ce{NH4NO3} $:
$ \ce{Li3N} + 3\ce{NH4NO3} -> \ce{LiNO3} + \ce{(NH4)3N} $

## Step 2: Balance $ \ce{NO3-} $ (from $ \ce{NH4NO3} $ and $ \ce{LiNO3} $)
On the left, $ 3\ce{NH4NO3} $ has 3 $ \ce{NO3-} $, so we put a coefficient of 3 in front of $ \ce{LiNO3} $:
$ \ce{Li3N} + 3\ce{NH4NO3} -> 3\ce{LiNO3} + \ce{(NH4)3N} $

## Step 3: Balance Li atoms
On the left, $ \ce{Li3N} $ has 3 Li, and on the right, $ 3\ce{LiNO3} $ has 3 Li – balanced.

## Step 4: Balance N atoms (from $ \ce{Li3N} $ and $ \ce{(NH4)3N} $)
Left: $ \ce{Li3N} $ has 1 N, $ 3\ce{NH4NO3} $ has 3 N (from $ \ce{NO3-} $) + 3 N (from $ \ce{NH4+} $)? Wait, no: $ \ce{NH4NO3} $ has 1 N in $ \ce{NH4+} $ and 1 N in $ \ce{NO3-} $, so 3$ \ce{NH4NO3} $ has 3 N (from $ \ce{NH4+} $) and 3 N (from $ \ce{NO3-} $). Right: $ \ce{(NH4)3N} $ has 1 N (from $ \ce{N^{3-}} $) and 3 N (from $ \ce{NH4+} $), and $ 3\ce{LiNO3} $ has 3 N (from $ \ce{NO3-} $). Wait, let’s check total N:
- Left: $ \ce{Li3N} $ (1 N) + $ 3\ce{NH4NO3} $ (3 N from $ \ce{NH4+} $ + 3 N from $ \ce{NO3-} $) = 7 N? Wait, no, maybe better to check all atoms:

Wait, $ \ce{Li3N} $: Li=3, N=1
$ 3\ce{NH4NO3} $: N=3 (from $ \ce{NH4+} $) + 3 (from $ \ce{NO3-} $) = 6 N; H=12; O=9
$ 3\ce{LiNO3} $: Li=3, N=3, O=9
$ \ce{(NH4)3N} $: N=1 (from $ \ce{N^{3-}} $) + 3 (from $ \ce{NH4+} $) = 4 N; H=12

Wait, N on left: 1 (from $ \ce{Li3N} $) + 6 (from $ 3\ce{NH4NO3} $) = 7 N. On right: 3 (from $ 3\ce{LiNO3} $) + 4 (from $ \ce{(NH4)3N} $) = 7 N – balanced.
Li: 3 on left, 3 on right – balanced.
H: 12 on left (3$ \ce{NH4NO3} $ has 3×4=12 H), 12 on right ($ \ce{(NH4)3N} $ has 3×4=12 H) – balanced.
O: 9 on left (3$ \ce{NH4NO3} $ has 3×3=9 O), 9 on right (3$ \ce{LiNO3} $ has 3×3=9 O) – balanced.

Wait, but the original problem’s handwritten answer has coefficients 1, 3, 3, 1. Let’s confirm:
$ 1\ce{Li3N} + 3\ce{NH4NO3} -> 3\ce{LiNO3} + 1\ce{(NH4)3N} $ – this balances all atoms.

## Step 5: Identify reaction type
This is a double replacement (DR) reaction because $ \ce{Li+} $ (from $ \ce{Li3N} $) and $ \ce{NH4+} $ (from $ \ce{NH4NO3} $) switch anions ($ \ce{N^{3-}} $ and $ \ce{NO3-} $).

# Answer:
Balanced equation: $ \boldsymbol{\ce{Li3N} + 3\ce{NH4NO3} -> 3\ce{LiNO3} + \ce{(NH4)3N}} $
Reaction type: $ \boldsymbol{DR} $

### Problem 5: Balancing $ \boldsymbol{\ce{HBr + Al(OH)3 -> H2O + AlBr3}} $ and Identifying Reaction Type
# Explanation:
## Step 1: Balance Br atoms
On the right, $ \ce{AlBr3} $ has 3 Br, so we put a coefficient of 3 in front of $ \ce{HBr} $:
$ 3\ce{HBr} + \ce{Al(OH)3} -> \ce{H2O} + \ce{AlBr3} $

## Step 2: Balance H and O atoms (from $ \ce{HBr} $, $ \ce{Al(OH)3} $, and $ \ce{H2O} $)
$ \ce{Al(OH)3} $ has 3 $ \ce{OH-} $, which will react with $ \ce{H+} $ from $ \ce{HBr} $ to form $ \ce{H2O} $. Each $ \ce{OH-} $ needs 1 $ \ce{H+} $ to make $ \ce{H2O} $, so 3 $ \ce{OH-} $ need 3 $ \ce{H+} $ (which we already have from $ 3\ce{HBr} $) to form 3 $ \ce{H2O} $. So put a coefficient of 3 in front of $ \ce{H2O} $:
$ 3\ce{HBr} + \ce{Al(OH)3} -> 3\ce{H2O} + \ce{AlBr3} $

## Step 3: Check Al atoms
1 Al on left ($ \ce{Al(OH)3} $) and 1 Al on right ($ \ce{AlBr3} $) – balanced.

## Step 4: Identify reaction type
This is a double replacement (DR) reaction (specifically a neutralization reaction, which is a type of DR) because $ \ce{H+} $ (from $ \ce{HBr} $) and $ \ce{Al^{3+}} $ (from $ \ce{Al(OH)3} $) switch anions ($ \ce{Br-} $ and $ \ce{OH-} $).

# Answer:
Balanced equation: $ \boldsymbol{3\ce{HBr} + \ce{Al(OH)3} -> 3\ce{H2O} + \ce{AlBr3}} $
Reaction type: $ \boldsymbol{DR} $

### Problem 6: Balancing $ \boldsymbol{\ce{C10H8 + O2 -> CO2 + H2O}} $ and Identifying Reaction Type
# Explanation:
## Step 1: Balance C atoms
$ \ce{C10H8} $ has 10 C, so put a coefficient of 10 in front of $ \ce{CO2} $:
$ \ce{C10H8} + \ce{O2} -> 10\ce{CO2} + \ce{H2O} $

## Step 2: Balance H atoms
$ \ce{C10H8} $ has 8 H, so put a coefficient of 4 in front of $ \ce{H2O} $ (since $ 4 \times 2 = 8 $ H):
$ \ce{C10H8} + \ce{O2} -> 10\ce{CO2} + 4\ce{H2O} $

## Step 3: Balance O atoms
On the right, $ 10\ce{CO2} $ has $ 10 \times 2 = 20 $ O, and $ 4\ce{H2O} $ has $ 4 \times 1 = 4 $ O – total 24 O. On the left, $ \ce{O2} $ has 2 O per molecule, so we need $ \frac{24}{2} = 12 $ O₂:
$ \ce{C10H8} + 12\ce{O2} -> 10\ce{CO2} + 4\ce{H2O} $

## Step 4: Identify reaction type
This is a combustion (C) reaction because a hydrocarbon ($ \ce{C10H8} $) reacts with $ \ce{O2} $ to form $ \ce{CO2} $ and $ \ce{H2O} $.

# Answer:
Balanced equation: $ \boldsymbol{\ce{C10H8} + 12\ce{O2} -> 10\ce{CO2} + 4\ce{H2O}} $
Reaction type: $ \boldsymbol{C} $

### Problem 7: Balancing $ \boldsymbol{\ce{Fe + S8 -> FeS}} $ and Identifying

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Explanation:

Problem 1: Balancing \( \boldsymbol{\ce{NaBr + Ca(OH)2 -> CaBr2 + NaOH}} \) and Identifying Reaction Type

Step 1: Balance Br atoms

Step 2: Balance Na atoms

Step 3: Check other atoms (Ca, O, H)

Step 4: Identify reaction type

Step 1: Balance \( \ce{NH4+} \) (from \( \ce{NH3} \) and \( \ce{(NH4)2SO4} \))

Step 2: Check other atoms (S, O, H)

Step 3: Identify reaction type

Step 1: Balance \( \ce{PO4^{3-}} \) (from \( \ce{H3PO4} \) and \( \ce{Pb3(PO4)2} \))

Step 2: Balance Pb atoms

Step 3: Balance H atoms

Step 4: Check O atoms

Step 5: Identify reaction type

Answer:

Problem 2: Balancing \( \boldsymbol{\ce{NH3 + H2SO4 -> (NH4)2SO4}} \) and Identifying Reaction Type