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. `1,3,1`
B. `4,1,4`
C. `3,2,3`
D. `2,1,2`

Explanation

To solve the given redox reaction and find the coefficients \( x \), \( y \), and \( z \) for the reaction:

\[ K_2Cr_2O_7 + x H_2SO_4 + y SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + z H_2O \]

we need to balance the reaction step by step.

### Step 1: Identify the oxidation states
- In \( K_2Cr_2O_7 \), chromium (Cr) has an oxidation state of +6.
- In \( Cr_2(SO_4)_3 \), chromium has an oxidation state of +3.
- In \( SO_2 \), sulfur (S) has an oxidation state of +4.
- In \( H_2SO_4 \) and \( K_2SO_4 \), sulfur has an oxidation state of +6.

### Step 2: Write the half-reactions
1. **Oxidation half-reaction**:
- \( SO_2 \) is oxidized to \( SO_4^{2-} \).
- The change in oxidation state is from +4 to +6, which involves the loss of 2 electrons.
- The half-reaction can be written as:
\[ SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^- \]

2. **Reduction half-reaction**:
- \( Cr_2O_7^{2-} \) is reduced to \( Cr^{3+} \).
- The change in oxidation state is from +6 to +3, which involves the gain of 6 electrons.
- The half-reaction can be written as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

### Step 3: Balance the electrons
To balance the electrons, we need to multiply the oxidation half-reaction by 3 (to equal the 6 electrons in the reduction half-reaction):
\[ 3(SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-) \]
This gives:
\[ 3SO_2 + 6H_2O \rightarrow 3SO_4^{2-} + 12H^+ + 6e^- \]

### Step 4: Combine the half-reactions
Now we can combine the balanced half-reactions:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- + 3SO_2 + 6H_2O \rightarrow 2Cr^{3+} + 7H_2O + 3SO_4^{2-} + 12H^+ + 6e^- \]

### Step 5: Simplify the equation
After canceling out the electrons and simplifying, we have:
\[ Cr_2O_7^{2-} + 3SO_2 + 2H_2O + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 7H_2O \]

### Step 6: Add potassium and sulfate
Now we need to include \( K_2Cr_2O_7 \) and \( H_2SO_4 \):
\[ K_2Cr_2O_7 + 3H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 1H_2O \]

### Final coefficients
From the balanced equation, we can now identify:
- \( x = 3 \) (coefficient of \( H_2SO_4 \))
- \( y = 3 \) (coefficient of \( SO_2 \))
- \( z = 1 \) (coefficient of \( H_2O \))

### Final Answer
Thus, the values of \( x \), \( y \), and \( z \) are:
- \( x = 3 \)
- \( y = 3 \)
- \( z = 1 \)