In the chemical reaction,
`K_(2)Cr_(2)O_(7)+xH_(2)SO_(4)+ySO_(2)rarrK_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+zH_(2)O`
`x,y, and z` are

Options

A. x=1,y=3,z-=1
B. x=4,y=1,z=4
C. x=3.y=2.z=1
D. x=2.y=2,z=1

Answer

To solve the given chemical reaction and find the values of \( x \), \( y \), and \( z \) in the equation:

\[ K_2Cr_2O_7 + xH_2SO_4 + ySO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + zH_2O \]

we will follow these steps:

### Step 1: Identify Oxidation States
- In \( K_2Cr_2O_7 \), the oxidation state of chromium (Cr) is +6.
- In \( Cr_2(SO_4)_3 \), the oxidation state of chromium is +3.
- In \( SO_2 \), the oxidation state of sulfur (S) is +4.
- In \( K_2SO_4 \), the oxidation state of sulfur is +6.

### Step 2: Determine Changes in Oxidation States
- The change for chromium from +6 to +3 indicates a reduction of 3 for each chromium atom. Since there are 2 chromium atoms, the total change is \( 2 \times 3 = 6 \) electrons gained.
- The change for sulfur from +4 in \( SO_2 \) to +6 in \( K_2SO_4 \) indicates an oxidation of 2 electrons for each sulfur atom.

### Step 3: Balance the Electrons
- To balance the electrons, we need to ensure that the total number of electrons lost equals the total number of electrons gained. Since 3 moles of \( SO_2 \) will provide \( 3 \times 2 = 6 \) electrons, we can write:

\[ \text{Reduction: } Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+} \]
\[ \text{Oxidation: } 3SO_2 \rightarrow 3SO_4^{2-} + 6e^- \]

### Step 4: Write the Balanced Reaction
- The balanced half-reactions can be combined:

\[ Cr_2O_7^{2-} + 3SO_2 \rightarrow 2Cr^{3+} + 3SO_4^{2-} \]

### Step 5: Include Other Components
- Now we need to include \( K_2SO_4 \) and \( H_2O \) in the balanced equation. The complete reaction is:

\[ K_2Cr_2O_7 + 3SO_2 + 2H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 2H_2O \]

### Step 6: Determine Values of \( x \), \( y \), and \( z \)
- From the balanced equation:
- \( x = 2 \) (for \( H_2SO_4 \))
- \( y = 3 \) (for \( SO_2 \))
- \( z = 2 \) (for \( H_2O \))

### Conclusion
The values of \( x \), \( y \), and \( z \) are:
- \( x = 2 \)
- \( y = 3 \)
- \( z = 2 \)

### Final Answer
Thus, the answer is \( x = 2, y = 3, z = 2 \).