To calculate the mass of oxygen produced from the decomposition of potassium chlorate (KClO3), we can follow these steps:
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Write down the balanced chemical equation: \[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \]
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Calculate the molar mass of potassium chlorate (KClO3):
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol
The molar mass of KClO3 is: \[ 39.10 , (\text{K}) + 35.45 , (\text{Cl}) + 3 \times 16.00 , (\text{O}) = 39.10 + 35.45 + 48.00 = 122.55 , \text{g/mol} \]
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Convert the mass of KClO3 to moles: Using the given mass of potassium chlorate: \[ \text{moles of KClO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{75.0 , \text{g}}{122.55 , \text{g/mol}} \approx 0.612 , \text{mol} \]
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Use the stoichiometry of the reaction to find moles of O2 produced: From the balanced equation, 2 moles of KClO3 produce 3 moles of O2. Thus, the ratio of KClO3 to O2 is: \[ \frac{3 , \text{mol O}_2}{2 , \text{mol KClO}_3} \]
Therefore, the moles of O2 produced from 0.612 moles of KClO3 is: \[ \text{moles of O}_2 = 0.612 , \text{mol KClO}_3 \times \frac{3 , \text{mol O}_2}{2 , \text{mol KClO}_3} = 0.918 , \text{mol O}_2 \]
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Calculate the mass of O2 produced: The molar mass of O2 (oxygen gas) is: \[ 2 \times 16.00 , \text{g/mol} = 32.00 , \text{g/mol} \]
Now, we can find the mass of O2 produced: \[ \text{mass of O}_2 = \text{moles of O}_2 \times \text{molar mass of O}_2 = 0.918 , \text{mol} \times 32.00 , \text{g/mol} \approx 29.376 , \text{g} \]
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Final answer: \[ \text{The mass of oxygen produced is approximately } 29.4 , \text{g}. \]
Thus, the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate is approximately 29.4 g.