In the redox reactin
`xKMnO_(4)+NH_(3)rarryKNO_(3)+MnO_(2)+KOH+H_(2)O`

Options

A. x=4,y=6
B. x=3,y=8
C. x=8,y=6
D. x=8,y=3

Explanation

To balance the redox reaction \( xKMnO_4 + NH_3 \rightarrow KNO_3 + MnO_2 + KOH + H_2O \), we will follow these steps:

### Step 1: Identify Oxidation States
First, we need to determine the oxidation states of the elements involved in the reaction.

- In \( KMnO_4 \), the oxidation state of Mn is +7.
- In \( NH_3 \), the oxidation state of N is -3.
- In \( KNO_3 \), the oxidation state of N is +5.
- In \( MnO_2 \), the oxidation state of Mn is +4.
- In \( KOH \), the oxidation state of K is +1 and O is -2.
- In \( H_2O \), the oxidation state of H is +1 and O is -2.

### Step 2: Determine Changes in Oxidation States
Next, we analyze the changes in oxidation states to identify which species are oxidized and reduced.

- **For Manganese**:
- From +7 in \( KMnO_4 \) to +4 in \( MnO_2 \) (Reduction).
- Change: \( +7 \) to \( +4 \) = 3 electrons gained.

- **For Nitrogen**:
- From -3 in \( NH_3 \) to +5 in \( KNO_3 \) (Oxidation).
- Change: \( -3 \) to \( +5 \) = 8 electrons lost.

### Step 3: Balance the Electrons
To balance the reaction, we need to ensure that the number of electrons lost equals the number of electrons gained.

- Since 8 electrons are lost and 3 electrons are gained, we can find a common multiple. The least common multiple of 3 and 8 is 24.
- Therefore, we will multiply the half-reactions accordingly:
- Multiply the reduction half-reaction by 8 (to account for 24 electrons gained).
- Multiply the oxidation half-reaction by 3 (to account for 24 electrons lost).

### Step 4: Write the Balanced Equation
Using the coefficients derived from balancing the electrons:

- For \( KMnO_4 \): \( 8 \times KMnO_4 \)
- For \( NH_3 \): \( 3 \times NH_3 \)

The balanced equation becomes:
\[ 8KMnO_4 + 3NH_3 \rightarrow 3KNO_3 + 8MnO_2 + 3KOH + 2H_2O \]

### Step 5: Final Check
Check if the number of atoms of each element and the charge are balanced on both sides of the equation.

- **Potassium (K)**: 8 on both sides.
- **Manganese (Mn)**: 8 on both sides.
- **Nitrogen (N)**: 3 on both sides.
- **Oxygen (O)**: 32 on both sides (8 from \( KMnO_4 \) and 3 from \( KNO_3 \) + 8 from \( MnO_2 \) + 3 from \( KOH \) + 2 from \( H_2O \)).
- **Hydrogen (H)**: 6 on both sides.

All elements are balanced.

### Final Answer
The balanced equation is:
\[ 8KMnO_4 + 3NH_3 \rightarrow 3KNO_3 + 8MnO_2 + 3KOH + 2H_2O \]