Complete the given equations:
(i) `Cu + 8 HNO_(3) rarr 3 Cu(NO_(3))_(2) + overset (W)......+ 4H_(2)O`
(ii) `4Zn + 10HNO_(3) rarr 4Zn(NO_(3))_(2) + 5H_(2)O + overset(X)....`
(iii) `l2 + 10 HNO_(3)rarr overset(Y)....+10 NO_(2) + 4H_(2)O `

Options

A. `{:("W","X","Y"),(2NO_(2),NO,5HIO_(3)):}`
B. `{:("W","X","Y"),(2NO,N_(2)O,2HIO_(3)):}`
C. `{:("W","X","Y"),(N_(2),NO_(2),HI):}`
D. `{:("W","X","Y"),(N_(2)O,NO_(2),3HI):}`

Explanation

To complete the given equations, we will analyze each reaction step by step and identify the missing products (W, X, and Y).

### Step 1: Completing the First Equation
**Given Reaction:**
\[ \text{Cu} + 8 \text{HNO}_3 \rightarrow 3 \text{Cu(NO}_3)_2 + W + 4 \text{H}_2\text{O} \]

**Balancing the Reaction:**
1. Copper (Cu) reacts with nitric acid (HNO3).
2. The products include copper(II) nitrate, water, and a nitrogen oxide.
3. The balanced equation shows that for every 3 moles of Cu(NO3)2 produced, 2 moles of NO are generated.

**Final Equation:**
\[ \text{Cu} + 8 \text{HNO}_3 \rightarrow 3 \text{Cu(NO}_3)_2 + 2 \text{NO} + 4 \text{H}_2\text{O} \]

**Thus, W = 2 NO.**

### Step 2: Completing the Second Equation
**Given Reaction:**
\[ 4 \text{Zn} + 10 \text{HNO}_3 \rightarrow 4 \text{Zn(NO}_3)_2 + 5 \text{H}_2\text{O} + X \]

**Balancing the Reaction:**
1. Zinc (Zn) reacts with nitric acid.
2. The products include zinc nitrate, water, and a nitrogen oxide.
3. The balanced equation indicates that the nitrogen oxide produced is nitrous oxide (N2O).

**Final Equation:**
\[ 4 \text{Zn} + 10 \text{HNO}_3 \rightarrow 4 \text{Zn(NO}_3)_2 + 5 \text{H}_2\text{O} + 2 \text{N}_2\text{O} \]

**Thus, X = 2 N2O.**

### Step 3: Completing the Third Equation
**Given Reaction:**
\[ \text{I}_2 + 10 \text{HNO}_3 \rightarrow Y + 10 \text{NO}_2 + 4 \text{H}_2\text{O} \]

**Balancing the Reaction:**
1. Iodine (I2) reacts with nitric acid.
2. The products include nitrogen dioxide and water.
3. The balanced equation indicates that the product is iodic acid (HIO3).

**Final Equation:**
\[ \text{I}_2 + 10 \text{HNO}_3 \rightarrow 4 \text{HIO}_3 + 10 \text{NO}_2 + 4 \text{H}_2\text{O} \]

**Thus, Y = 4 HIO3.**

### Summary of Results
- W = 2 NO
- X = 2 N2O
- Y = 4 HIO3