How many grams of potassium chlorate should be decomposed to liberate `9.6` g of oxygen?
Equation: `2KClO_(3) to 2KCl + 3O_(2)`
Relative atomic mass:`K = 39, Cl = 35.5, O = 16`

Explanation

To solve the problem of how many grams of potassium chlorate (KClO₃) should be decomposed to liberate 9.6 g of oxygen (O₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of potassium chlorate can be represented by the following balanced equation:
\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]

### Step 2: Determine the molar mass of potassium chlorate (KClO₃)
To find the molar mass of KClO₃, we add the atomic masses of its constituent elements:
- Potassium (K) = 39 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Oxygen (O) = 16 g/mol (and there are 3 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of KClO}_3 = 39 + 35.5 + (3 \times 16) = 39 + 35.5 + 48 = 122.5 \text{ g/mol} \]

### Step 3: Calculate the mass of oxygen produced from the decomposition
From the balanced equation, we see that 2 moles of KClO₃ produce 3 moles of O₂.

First, we need to find out how many moles of O₂ correspond to 9.6 g:
\[ \text{Molar mass of O}_2 = 2 \times 16 = 32 \text{ g/mol} \]
\[ \text{Moles of O}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{9.6 \text{ g}}{32 \text{ g/mol}} = 0.3 \text{ moles} \]

### Step 4: Relate moles of O₂ to moles of KClO₃
From the balanced equation:
- 3 moles of O₂ are produced from 2 moles of KClO₃.

Using the mole ratio:
\[ \frac{2 \text{ moles KClO}_3}{3 \text{ moles O}_2} = \frac{x \text{ moles KClO}_3}{0.3 \text{ moles O}_2} \]
Cross-multiplying gives:
\[ x = \frac{2}{3} \times 0.3 = 0.2 \text{ moles of KClO}_3 \]

### Step 5: Calculate the mass of KClO₃ required
Now, we can find the mass of KClO₃ needed using its molar mass:
\[ \text{Mass of KClO}_3 = \text{moles} \times \text{molar mass} = 0.2 \text{ moles} \times 122.5 \text{ g/mol} = 24.5 \text{ g} \]

### Final Answer
Thus, the mass of potassium chlorate that should be decomposed to liberate 9.6 g of oxygen is approximately **24.5 g**.

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