When heated, potassium permangnate decomposes according to the following equation :
`2KMnO_(4)toK_(2)MnO_(4)+MnO_(2)+O_(2)`
Given that the molecular mass of potassium permangnate is 158, what volume of oxygen measured at room temperature, would be obtained by the complete decomposition of 15.8 g of potassium permangnate? (Molar volume of room temperature is 22.4 litres)

Explanation

To solve the problem step-by-step, we will follow these instructions:

### Step 1: Write down the decomposition reaction
The decomposition of potassium permanganate (KMnO₄) is represented by the following equation:
\[ 2 \text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2 \]

### Step 2: Determine the molar mass of KMnO₄
The problem states that the molecular mass of potassium permanganate (KMnO₄) is 158 g/mol.
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