In the redox reaction
`xMnO+yPbO_(2)+zHNO_(3) rarr HMnO_(4)+Pb(NO_(3))_(2)+H_(2)O`

Options

A. `x=2`, `y=5`, `z=10`
B. `x=2`, `y=7`, `z=8`
C. `x=2`, `y=5`, `z=8`
D. `x=2`, `y=5`, `z=5`

Explanation

To solve the redox reaction given by the equation:

\[ x \text{MnO} + y \text{PbO}_2 + z \text{HNO}_3 \rightarrow \text{HMnO}_4 + \text{Pb(NO}_3)_2 + \text{H}_2\text{O} \]

we need to balance the reaction by determining the values of \( x \), \( y \), and \( z \).

### Step 1: Identify the Oxidation and Reduction Half-Reactions

1. **Oxidation Half-Reaction**:
- Manganese in MnO is oxidized to MnO₄.
- The half-reaction is:
\[ \text{MnO} \rightarrow \text{HMnO}_4 \]

2. **Reduction Half-Reaction**:
- Lead in PbO₂ is reduced to Pb(NO₃)₂.
- The half-reaction is:
\[ \text{PbO}_2 + \text{HNO}_3 \rightarrow \text{Pb(NO}_3)_2 \]

### Step 2: Balance the Oxidation Half-Reaction

1. **Balance Oxygen**:
- MnO has 1 oxygen, and HMnO₄ has 4 oxygen.
- Add 3 water molecules to the right:
\[ \text{MnO} + 3 \text{H}_2\text{O} \rightarrow \text{HMnO}_4 \]

2. **Balance Hydrogen**:
- Now we have 6 H on the right from 3 H₂O.
- Add 5 H⁺ to the left:
\[ \text{MnO} + 3 \text{H}_2\text{O} \rightarrow \text{HMnO}_4 + 5 \text{H}^+ \]

3. **Balance Charge**:
- The left side has a charge of +5 from 5 H⁺.
- Add 5 electrons to the left:
\[ \text{MnO} + 3 \text{H}_2\text{O} \rightarrow \text{HMnO}_4 + 5 \text{H}^+ + 5 \text{e}^- \]

### Step 3: Balance the Reduction Half-Reaction

1. **Balance Lead Compounds**:
- Start with:
\[ \text{PbO}_2 + \text{HNO}_3 \rightarrow \text{Pb(NO}_3)_2 \]
- To balance the nitrate, we need 2 HNO₃:
\[ \text{PbO}_2 + 2 \text{HNO}_3 \rightarrow \text{Pb(NO}_3)_2 \]

2. **Balance Oxygen**:
- On the left, we have 2 O from PbO₂ and 6 O from 2 HNO₃ (total 8 O).
- Add 2 H₂O to the left:
\[ \text{PbO}_2 + 2 \text{HNO}_3 + 2 \text{H}_2\text{O} \rightarrow \text{Pb(NO}_3)_2 \]

3. **Balance Hydrogen**:
- Now we have 4 H from 2 H₂O and 2 H from 2 HNO₃ (total 6 H).
- Add 2 H⁺ to the right:
\[ \text{PbO}_2 + 2 \text{HNO}_3 + 2 \text{H}_2\text{O} \rightarrow \text{Pb(NO}_3)_2 + 2 \text{H}^+ \]

4. **Balance Charge**:
- The left side has no charge, and the right side has a charge of +2 from 2 H⁺.
- Add 2 electrons to the right:
\[ \text{PbO}_2 + 2 \text{HNO}_3 + 2 \text{H}_2\text{O} \rightarrow \text{Pb(NO}_3)_2 + 2 \text{H}^+ + 2 \text{e}^- \]

### Step 4: Combine the Half-Reactions

1. **Multiply the Half-Reactions**:
- Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 5 to equalize the number of electrons:
\[ 2 \left( \text{MnO} + 3 \text{H}_2\text{O} \rightarrow \text{HMnO}_4 + 5 \text{H}^+ + 5 \text{e}^- \right) \]
\[ 5 \left( \text{PbO}_2 + 2 \text{HNO}_3 + 2 \text{H}_2\text{O} \rightarrow \text{Pb(NO}_3)_2 + 2 \text{H}^+ + 2 \text{e}^- \right) \]

2. **Final Balanced Reaction**:
- Combine the two balanced half-reactions:
\[ 2 \text{MnO} + 5 \text{PbO}_2 + 10 \text{HNO}_3 \rightarrow 2 \text{HMnO}_4 + 5 \text{Pb(NO}_3)_2 + 4 \text{H}_2\text{O} \]

### Step 5: Determine the Values of \( x \), \( y \), and \( z \)

From the balanced equation:
- \( x = 2 \) (coefficient of MnO)
- \( y = 5 \) (coefficient of PbO₂)
- \( z = 10 \) (coefficient of HNO₃)

### Final Answer:
The values are:
- \( x = 2 \)
- \( y = 5 \)
- \( z = 10 \)