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.
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Convert kg to g: \[ 6.0 , \text{kg} = 6.0 \times 1000 , \text{g} = 6000 , \text{g} \]
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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.
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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: \[ \text{Molar mass of KClO}_3 = 39.10 + 35.45 + (3 \times 16.00) = 122.55 , \text{g/mol} \]
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Convert grams of KClO₃ to moles: \[ \text{Moles of KClO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{6000 , \text{g}}{122.55 , \text{g/mol}} \approx 49.0 , \text{mol} \]
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Use the stoichiometry to find moles of O₂ produced: According to the balanced equation, 2 moles of KClO₃ yield 3 moles of O₂: \[ \text{Moles of O}_2 = \left(49.0 , \text{mol KClO}_3 \right) \times \frac{3 , \text{mol O}_2}{2 , \text{mol KClO}_3} = 73.5 , \text{mol O}_2 \]
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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).