Write the balanced equations for the reactions of the following compounds with water:
(i) `AI_(4)C_(3)` (ii) `CaNCN` (iii) `BF_(3)` (iv) `NCI_(3)` (v) `XeF_(4)`

Answer

To write the balanced equations for the reactions of the given compounds with water, we will analyze each compound step by step.

### (i) Reaction of Aluminum Carbide (Al₄C₃) with Water

1. **Write the unbalanced equation:**
\[ \text{Al}_4\text{C}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{CH}_4 \]

2. **Balance the equation:**
- Aluminum (Al): 4 Al on the left, so we need 4 Al(OH)₃ on the right.
- Carbon (C): 3 C on the left, so we need 3 CH₄ on the right.
- Hydrogen (H): 4 (from Al(OH)₃) + 12 (from CH₄) = 16 H on the right. We need 8 H₂O on the left.
- Oxygen (O): 8 O from 8 H₂O = 8 O on the left, and 4 O from 4 Al(OH)₃ on the right.

The balanced equation is:
\[ \text{Al}_4\text{C}_3 + 12 \text{H}_2\text{O} \rightarrow 4 \text{Al(OH)}_3 + 3 \text{CH}_4 \]

### (ii) Reaction of Calcium Cyanamide (CaNCN) with Water

1. **Write the unbalanced equation:**
\[ \text{CaNCN} + \text{H}_2\text{O} \rightarrow \text{CaCO}_3 + \text{NH}_4\text{OH} \]

2. **Balance the equation:**
- Calcium (Ca): 1 on both sides.
- Nitrogen (N): 2 N on the left, so we need 2 NH₄OH on the right.
- Carbon (C): 1 C on both sides.
- Oxygen (O): 3 O from CaCO₃ and 2 O from 2 NH₄OH = 5 O on the right, so we need 5 H₂O on the left.

The balanced equation is:
\[ \text{CaNCN} + 5 \text{H}_2\text{O} \rightarrow \text{CaCO}_3 + 2 \text{NH}_4\text{OH} \]

### (iii) Reaction of Boron Trifluoride (BF₃) with Water

1. **Write the unbalanced equation:**
\[ \text{BF}_3 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + \text{HBF}_4 \]

2. **Balance the equation:**
- Boron (B): 1 B on the left, so we need 1 H₃BO₃ on the right.
- Fluorine (F): 3 F from 3 HBF₄ on the right, so we need 3 BF₃ on the left.
- Hydrogen (H): 3 H from H₃BO₃ and 3 H from 3 HBF₄ = 6 H on the right, so we need 3 H₂O on the left.

The balanced equation is:
\[ 4 \text{BF}_3 + 3 \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + 3 \text{HBF}_4 \]

### (iv) Reaction of Nitrogen Trichloride (NCl₃) with Water

1. **Write the unbalanced equation:**
\[ \text{NCl}_3 + \text{H}_2\text{O} \rightarrow \text{NH}_3 + \text{HOCl} \]

2. **Balance the equation:**
- Nitrogen (N): 1 N on both sides.
- Chlorine (Cl): 3 Cl on the left, so we need 3 HOCl on the right.
- Hydrogen (H): 3 H from NH₃ and 3 H from 3 HOCl = 6 H on the right, so we need 3 H₂O on the left.

The balanced equation is:
\[ \text{NCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{NH}_3 + 3 \text{HOCl} \]

### (v) Reaction of Xenon Tetrafluoride (XeF₄) with Water

1. **Write the unbalanced equation:**
\[ \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + \text{HF} \]

2. **Balance the equation:**
- Xenon (Xe): 1 Xe on the left, so we need 1 Xe on the right.
- Fluorine (F): 4 F on the left, so we need 4 HF on the right.
- Oxygen (O): 3 O from XeO₃ on the right, so we need 12 H₂O on the left.

The balanced equation is:
\[ \text{XeF}_4 + 2 \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + 4 \text{HF} \]

### Summary of Balanced Equations
1. \( \text{Al}_4\text{C}_3 + 12 \text{H}_2\text{O} \rightarrow 4 \text{Al(OH)}_3 + 3 \text{CH}_4 \)
2. \( \text{CaNCN} + 5 \text{H}_2\text{O} \rightarrow \text{CaCO}_3 + 2 \text{NH}_4\text{OH} \)
3. \( 4 \text{BF}_3 + 3 \text{H}_2\text{O} \rightarrow \text{H}_3\text{BO}_3 + 3 \text{HBF}_4 \)
4. \( \text{NCl}_3 + 3 \text{H}_2\text{O} \rightarrow \text{NH}_3 + 3 \text{HOCl} \)
5. \( \text{XeF}_4 + 2 \text{H}_2\text{O} \rightarrow \text{Xe} + \text{XeO}_3 + 4 \text{HF} \)