By starting with 0.5 moles of sodium peroxide how many moles of dioxygen gas can be obtained by dropping excess of acidified potassium permanganate solution on it ?

Options

A. 0.125 mole
B. 1 mole
C. 0.25 mole
D. 0.5 mole

Answer

To determine how many moles of dioxygen gas (O2) can be obtained from 0.5 moles of sodium peroxide (Na2O2) when reacted with excess acidified potassium permanganate (KMnO4), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between sodium peroxide and acidified potassium permanganate can be represented as follows:
\[ 2 \text{KMnO}_4 + 5 \text{Na}_2\text{O}_2 + 8 \text{H}_2\text{SO}_4 \rightarrow 2 \text{MnSO}_4 + 5 \text{O}_2 + 8 \text{Na}_2\text{SO}_4 + 8 \text{H}_2\text{O} \]

2. **Determine the Stoichiometry**:
From the balanced equation, we can see that 5 moles of sodium peroxide (Na2O2) produce 5 moles of dioxygen (O2). Therefore, the stoichiometric ratio is:
\[ 5 \text{Na}_2\text{O}_2 \rightarrow 5 \text{O}_2 \]
This means that 1 mole of Na2O2 produces 1 mole of O2.

3. **Calculate the Moles of O2 from Na2O2**:
Since we have 0.5 moles of sodium peroxide (Na2O2), we can use the stoichiometric ratio to find the moles of O2 produced:
\[ \text{Moles of } O_2 = 0.5 \text{ moles of } Na_2O_2 \times \frac{1 \text{ mole of } O_2}{1 \text{ mole of } Na_2O_2} = 0.5 \text{ moles of } O_2 \]

4. **Conclusion**:
Therefore, from 0.5 moles of sodium peroxide, we can obtain **0.5 moles of dioxygen gas (O2)**.

### Final Answer:
0.5 moles of dioxygen gas (O2) can be obtained from 0.5 moles of sodium peroxide.

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