`2KMnO_(4)+H_(2)C_(2)O_(4)toK_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O`. ltvrgt Potassium permanganate oxidises oxalic acid in acidic medium as per the above chemical reaction. Identify the coefficient of `H_(2)SO_(4)` in the above equatio.

Options

A. 3
B. 4
C. 5
D. 2

Explanation

To identify the coefficient of \( H_2SO_4 \) in the given chemical reaction, we need to balance the equation step by step. The reaction is:

\[ 2KMnO_4 + H_2C_2O_4 + H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O \]

### Step 1: Write the unbalanced equation
The unbalanced equation is:

\[ 2KMnO_4 + H_2C_2O_4 + H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O \]

### Step 2: Count the number of atoms of each element on both sides
- **Reactants:**
- K: 2
- Mn: 2
- C: 2
- H: 2 (from \( H_2C_2O_4 \)) + 2 (from \( H_2SO_4 \)) = 4
- O: 8 (from \( 2KMnO_4 \)) + 4 (from \( H_2C_2O_4 \)) + 4 (from \( H_2SO_4 \)) = 16

- **Products:**
- K: 2 (from \( K_2SO_4 \))
- Mn: 2 (from \( 2MnSO_4 \))
- C: 10 (from \( 10CO_2 \))
- H: 16 (from \( 8H_2O \))
- O: 4 (from \( K_2SO_4 \)) + 8 (from \( 2MnSO_4 \)) + 20 (from \( 10CO_2 \)) + 8 (from \( 8H_2O \)) = 40

### Step 3: Balance the carbon atoms
Since there are 10 carbon atoms on the product side (from \( 10CO_2 \)), we need to adjust the oxalic acid (\( H_2C_2O_4 \)):

\[ 2KMnO_4 + 5H_2C_2O_4 + H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O \]

### Step 4: Count the atoms again
Now recount the atoms:
- **Reactants:**
- K: 2
- Mn: 2
- C: 10 (from \( 5H_2C_2O_4 \))
- H: 10 (from \( 5H_2C_2O_4 \)) + 2 (from \( H_2SO_4 \)) = 12
- O: 16 (from \( 2KMnO_4 \)) + 10 (from \( 5H_2C_2O_4 \)) + 4 (from \( H_2SO_4 \)) = 30

- **Products:**
- K: 2
- Mn: 2
- C: 10
- H: 16
- O: 4 + 8 + 20 + 8 = 40

### Step 5: Balance the sulfur atoms
To balance the sulfur atoms, we need to add coefficients to \( H_2SO_4 \):

\[ 2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O \]

### Step 6: Final count of atoms
Now recount the atoms:
- **Reactants:**
- K: 2
- Mn: 2
- C: 10
- H: 10 + 6 = 16
- O: 16 + 10 + 12 = 38

- **Products:**
- K: 2
- Mn: 2
- C: 10
- H: 16
- O: 4 + 8 + 20 + 12 = 44

### Conclusion
After balancing, we find that the coefficient of \( H_2SO_4 \) is **3**.