Consider the equations given below and find the sum of x, y and z.
(i) `Mg+underset("(very dilute)")(xHNO_(3))(5%)rarrMg(NO_(3))_(2)+H_(2)`
(ii) `Cu+underset("(dil.)")(yHNO_(3))rarr 3Cu(NO_(3))_(2)+2NO+4H_(2)O`
(iii) `I_(2)+zHNO_(3)rarr 2HIO_(3)+10NO_(2)+4H_(2)O`

Answer

To find the sum of x, y, and z from the given chemical equations, we will balance each equation step by step.

### Step 1: Balancing the first equation
The first equation is:
\[ \text{Mg} + x \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2 \]

1. **Identify the reactants and products**:
- Reactants: Magnesium (Mg) and dilute nitric acid (HNO₃).
- Products: Magnesium nitrate (Mg(NO₃)₂) and hydrogen gas (H₂).

2. **Balancing the equation**:
- On the right side, we have 1 Mg, 2 nitrate groups (NO₃), and 1 H₂ molecule.
- To balance the nitrate groups, we need 2 HNO₃ on the left side.
- The balanced equation becomes:
\[ \text{Mg} + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2 \]
- Therefore, \( x = 2 \).

### Step 2: Balancing the second equation
The second equation is:
\[ \text{Cu} + y \text{HNO}_3 \rightarrow 3 \text{Cu(NO}_3\text{)}_2 + 2 \text{NO} + 4 \text{H}_2\text{O} \]

1. **Identify the reactants and products**:
- Reactants: Copper (Cu) and dilute nitric acid (HNO₃).
- Products: Copper nitrate (Cu(NO₃)₂), nitrogen monoxide (NO), and water (H₂O).

2. **Balancing the equation**:
- On the right side, we have 3 Cu, 6 nitrate groups (from 3 Cu(NO₃)₂), 2 NO, and 4 H₂O.
- To balance the nitrogen and hydrogen, we need to calculate:
- Total nitrogen on the right: \( 6 \) (from 6 NO₃) + \( 2 \) (from 2 NO) = \( 8 \).
- Total hydrogen from water: \( 4 \) (from 4 H₂O).
- Therefore, we need \( y = 8 \) to balance the equation:
\[ \text{Cu} + 8 \text{HNO}_3 \rightarrow 3 \text{Cu(NO}_3\text{)}_2 + 2 \text{NO} + 4 \text{H}_2\text{O} \]

### Step 3: Balancing the third equation
The third equation is:
\[ \text{I}_2 + z \text{HNO}_3 \rightarrow 2 \text{HIO}_3 + 10 \text{NO}_2 + 4 \text{H}_2\text{O} \]

1. **Identify the reactants and products**:
- Reactants: Iodine (I₂) and dilute nitric acid (HNO₃).
- Products: Iodic acid (HIO₃), nitrogen dioxide (NO₂), and water (H₂O).

2. **Balancing the equation**:
- On the right side, we have 2 I, 10 NO₂, and 4 H₂O.
- Total nitrogen from NO₂: \( 10 \).
- Total oxygen from products: \( 20 \) (from 10 NO₂) + \( 6 \) (from 2 HIO₃) + \( 4 \) (from 4 H₂O) = \( 30 \).
- Total hydrogen: \( 8 \) (from 2 HIO₃) + \( 4 \) (from 4 H₂O) = \( 10 \).
- To balance the equation, we need \( z = 10 \):
\[ \text{I}_2 + 10 \text{HNO}_3 \rightarrow 2 \text{HIO}_3 + 10 \text{NO}_2 + 4 \text{H}_2\text{O} \]

### Step 4: Calculating the sum of x, y, and z
Now that we have the values:
- \( x = 2 \)
- \( y = 8 \)
- \( z = 10 \)

The sum is:
\[ x + y + z = 2 + 8 + 10 = 20 \]

### Final Answer
The sum of \( x, y, \) and \( z \) is \( 20 \).

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