For the redox reaction given, what is the value of `(x)/(z)` ?
`xNO_(3)^(-)+yAs_(2)S_(3)+zH_(2)O to -----AsO_(4)^(3-)+-----NO+------SO_(4)^(2-)+-----H^(+)`.

Answer

To solve the problem, we need to balance the given redox reaction and find the ratio \( \frac{x}{z} \) where \( x \), \( y \), and \( z \) are the coefficients of \( NO_3^- \), \( As_2S_3 \), and \( H_2O \) respectively.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reaction is given as:
\[ x \, NO_3^- + y \, As_2S_3 + z \, H_2O \rightarrow \text{products} \]
The products are:
\[ AsO_4^{3-} + NO + SO_4^{2-} + H^+ \]

2. **Determine Oxidation States**:
- For \( NO_3^- \), the oxidation state of nitrogen (N) is +5.
- For \( NO \), the oxidation state of nitrogen (N) is +2.
- For \( As_2S_3 \), the oxidation state of arsenic (As) is -3.
- For \( AsO_4^{3-} \), the oxidation state of arsenic (As) is +5.
- For \( SO_4^{2-} \), the oxidation state of sulfur (S) is +6.

3. **Identify Oxidation and Reduction**:
- \( NO_3^- \) is reduced to \( NO \) (from +5 to +2).
- \( As_2S_3 \) is oxidized to \( AsO_4^{3-} \) (from -3 to +5).
- \( S \) in \( As_2S_3 \) is oxidized to \( SO_4^{2-} \) (from -2 to +6).

4. **Balance the Half-Reactions**:
- **Reduction half-reaction**:
\[ NO_3^- + 3e^- \rightarrow NO \]
This indicates that 1 mole of \( NO_3^- \) is reduced by gaining 3 electrons.

- **Oxidation half-reaction** for \( As_2S_3 \):
- For \( As \):
\[ As_2S_3 \rightarrow 2AsO_4^{3-} + 12e^- \]
- For \( S \):
\[ S \rightarrow SO_4^{2-} + 6e^- \]

5. **Combine the Half-Reactions**:
To balance the electrons, we need to ensure the total electrons lost equals the total electrons gained.
- The total electrons from \( As_2S_3 \) oxidation is 12 (from As) and 6 (from S), totaling 18 electrons.
- To balance, we multiply the reduction half-reaction by 6:
\[ 6 \, NO_3^- + 18e^- \rightarrow 6 \, NO \]

6. **Final Balancing**:
We can now balance the entire reaction:
\[ 6 \, NO_3^- + 3 \, As_2S_3 + 6 \, H_2O \rightarrow 6 \, NO + 6 \, SO_4^{2-} + 6 \, AsO_4^{3-} + 6 \, H^+ \]

7. **Identify Coefficients**:
From the balanced equation:
- \( x = 6 \) (for \( NO_3^- \))
- \( y = 3 \) (for \( As_2S_3 \))
- \( z = 6 \) (for \( H_2O \))

8. **Calculate \( \frac{x}{z} \)**:
\[ \frac{x}{z} = \frac{6}{6} = 1 \]

### Final Answer:
The value of \( \frac{x}{z} \) is \( 1 \).