Question

How many grams of oxygen can be obtained from 6.0 kg of KClO₃?
*
Convert kg to g

Answer 1

To determine how many grams of oxygen can be obtained from 6.0 kg of KClO₃ (potassium chlorate), we first need to convert the mass of KClO₃ from kilograms to grams.

1. **Convert kg to g:**
\[
6.0 \, \text{kg} = 6.0 \times 1000 \, \text{g} = 6000 \, \text{g}
\]

2. **Determine the balanced chemical equation:**
The decomposition of KClO₃ can be represented by the following balanced chemical reaction:
\[
2 \, \text{KClO}_3 \rightarrow 2 \, \text{KCl} + 3 \, \text{O}_2
\]
This equation states that 2 moles of KClO₃ produce 3 moles of O₂.

3. **Calculate the molar mass of KClO₃:**
- K (Potassium): 39.10 g/mol
- Cl (Chlorine): 35.45 g/mol
- O (Oxygen): 16.00 g/mol

So, the molar mass of KClO₃ is:
\[
\text{Molar mass of KClO}_3 = 39.10 + 35.45 + (3 \times 16.00) = 122.55 \, \text{g/mol}
\]

4. **Convert grams of KClO₃ to moles:**
\[
\text{Moles of KClO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{6000 \, \text{g}}{122.55 \, \text{g/mol}} \approx 49.0 \, \text{mol}
\]

5. **Use the stoichiometry to find moles of O₂ produced:**
According to the balanced equation, 2 moles of KClO₃ yield 3 moles of O₂:
\[
\text{Moles of O}_2 = \left(49.0 \, \text{mol KClO}_3 \right) \times \frac{3 \, \text{mol O}_2}{2 \, \text{mol KClO}_3} = 73.5 \, \text{mol O}_2
\]

6. **Convert moles of O₂ to grams:**
The molar mass of O₂ (Oxygen) is:
\[
\text{Molar mass of O}_2 = 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}
\]
Now we can find the mass of O₂ produced:
\[
\text{Mass of O}_2 = 73.5 \, \text{mol O}_2 \times 32.00 \, \text{g/mol} = 2352 \, \text{g}
\]

So, from 6.0 kg of KClO₃, you can obtain **2352 grams of oxygen (O₂)**.