In the balanced equation
`FeS+MnO_(4)^(-)overset(H^(+))rarrFe^(3+)+SO_(4)^(2-)+Mn^(2+)`
the stoichiometric coefficients of `FeS` and `MnO_(4)^(-)` are in the ratio `:`

Options

A. `8:5`
B. `5:8`
C. `9:5`
D. `5:9`

Answer

To find the stoichiometric coefficients of `FeS` and `MnO4^(-)` in the given balanced equation, we will follow these steps:

### Step 1: Write the Unbalanced Equation
The unbalanced equation is:
\[ \text{FeS} + \text{MnO}_4^{-} + \text{H}^{+} \rightarrow \text{Fe}^{3+} + \text{SO}_4^{2-} + \text{Mn}^{2+} \]

### Step 2: Determine the Oxidation States
- For `FeS`:
- Iron (Fe) is in the +2 oxidation state.
- Sulfur (S) is in the -2 oxidation state.

- For products:
- Iron (Fe) is in the +3 oxidation state.
- Sulfur (S) in sulfate (SO4^2-) is in the +6 oxidation state.

- For `MnO4^-`:
- Manganese (Mn) is in the +7 oxidation state.

- For products:
- Manganese (Mn) is in the +2 oxidation state.

### Step 3: Identify Oxidation and Reduction
- **Oxidation**:
- Iron (Fe) goes from +2 to +3 (oxidation).
- Sulfur (S) goes from -2 to +6 (oxidation).

- **Reduction**:
- Manganese (Mn) goes from +7 to +2 (reduction).

### Step 4: Calculate Changes in Oxidation States
- Total increase in oxidation state (oxidation):
- Fe: +1 (from +2 to +3)
- S: +8 (from -2 to +6)
- Total increase = 1 + 8 = 9

- Total decrease in oxidation state (reduction):
- Mn: -5 (from +7 to +2)

### Step 5: Balance the Equation
To balance the changes in oxidation states:
- The total increase in oxidation states (9) must equal the total decrease (5).
- To balance, we can use coefficients:
- For `FeS`, we use a coefficient of 5.
- For `MnO4^-`, we use a coefficient of 9.

Thus, the balanced equation becomes:
\[ 5 \text{FeS} + 9 \text{MnO}_4^{-} + \text{H}^{+} \rightarrow 5 \text{Fe}^{3+} + 5 \text{SO}_4^{2-} + 9 \text{Mn}^{2+} \]

### Step 6: Determine the Stoichiometric Coefficients Ratio
The stoichiometric coefficients of `FeS` and `MnO4^-` are 5 and 9, respectively. Therefore, the ratio is:
\[ \text{Ratio} = \frac{\text{Coefficient of FeS}}{\text{Coefficient of MnO}_4^{-}} = \frac{5}{9} \]

### Final Answer
The stoichiometric coefficients of `FeS` and `MnO4^(-)` are in the ratio **5:9**.

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