`NO_(2)^(-)` ion in`KNO_(2)`is oxidised to `NO_(3)^(-)` ion by the action of `KMnO_(4)` in `H_(2)SO_(4)` solution according to the reaction :
`5KNO_(2) + 2KMnO_(4) + 3H_(2)SO_(4) to 5KNO_(3) + 2MnSO_(4) + 5I_(2) + 8H_(2)O`
How much KMn04 are needed to oxidise 11.4 g of `KNO_(2)`?

Explanation

To determine how much KMnO4 is needed to oxidize 11.4 g of KNO2, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 5 \text{KNO}_2 + 2 \text{KMnO}_4 + 3 \text{H}_2\text{SO}_4 \rightarrow 5 \text{KNO}_3 + 2 \text{MnSO}_4 + 5 \text{I}_2 + 8 \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of KNO2
The molar mass of KNO2 can be calculated as follows:
- K (Potassium) = 39.1 g/mol
- N (Nitrogen) = 14.0 g/mol
- O (Oxygen) = 16.0 g/mol × 2 = 32.0 g/mol

Adding these together:
\[ \text{Molar mass of KNO}_2 = 39.1 + 14.0 + 32.0 = 85.1 \text{ g/mol} \]

### Step 3: Calculate the number of moles of KNO2 in 11.4 g
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
We can find the number of moles of KNO2:
\[ \text{Number of moles of KNO}_2 = \frac{11.4 \text{ g}}{85.1 \text{ g/mol}} \approx 0.134 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of KMnO4 needed
From the balanced equation, we see that 5 moles of KNO2 react with 2 moles of KMnO4. Therefore, the ratio of KNO2 to KMnO4 is:
\[ \frac{2 \text{ moles KMnO}_4}{5 \text{ moles KNO}_2} \]

Now, we can find the moles of KMnO4 needed for 0.134 moles of KNO2:
\[ \text{Moles of KMnO}_4 = 0.134 \text{ moles KNO}_2 \times \frac{2 \text{ moles KMnO}_4}{5 \text{ moles KNO}_2} \approx 0.054 \text{ moles KMnO}_4 \]

### Step 5: Calculate the molar mass of KMnO4
The molar mass of KMnO4 can be calculated as follows:
- K (Potassium) = 39.1 g/mol
- Mn (Manganese) = 54.9 g/mol
- O (Oxygen) = 16.0 g/mol × 4 = 64.0 g/mol

Adding these together:
\[ \text{Molar mass of KMnO}_4 = 39.1 + 54.9 + 64.0 = 158.0 \text{ g/mol} \]

### Step 6: Calculate the mass of KMnO4 needed
Using the number of moles and the molar mass, we can find the mass of KMnO4 needed:
\[ \text{Mass of KMnO}_4 = \text{moles} \times \text{molar mass} \]
\[ \text{Mass of KMnO}_4 = 0.054 \text{ moles} \times 158.0 \text{ g/mol} \approx 8.53 \text{ g} \]

### Final Answer
Therefore, approximately **8.53 g** of KMnO4 is needed to oxidize 11.4 g of KNO2.

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