To determine how many pounds of potassium chlorate (KClO₃) are needed to obtain 245 pounds of oxygen (O₂), we first need to look at the chemical reaction involved in the decomposition of potassium chlorate:
\[ 2 , \text{KClO}_3(s) \rightarrow 2 , \text{KCl}(s) + 3 , \text{O}_2(g) \]
From the reaction, we can see that:
- 2 moles of KClO₃ produce 3 moles of O₂.
Next, we need to calculate the molar masses to relate the amount of KClO₃ to the amount of oxygen produced:
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The molar mass of KClO₃ is:
- K: 39.10 g/mol
- Cl: 35.45 g/mol
- O₃: 3 × 16.00 g/mol = 48.00 g/mol
- Total = 39.10 + 35.45 + 48.00 = 122.55 g/mol
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The molar mass of O₂ is:
- O₂: 2 × 16.00 g/mol = 32.00 g/mol
Now, from the stoichiometry of the reaction, we can set up the proportion:
\[ \frac{2 , \text{moles KClO}_3}{3 , \text{moles O}_2} \]
We want to find how many moles of KClO₃ are needed to produce 245 pounds of O₂.
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Convert pounds of O₂ to grams: \[ 245 , \text{lb} \times 453.592 , \text{g/lb} = 111,130.84 , \text{g} \]
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Calculate the number of moles of O₂ produced: \[ \text{Moles of O}_2 = \frac{111,130.84 , \text{g}}{32.00 , \text{g/mol}} \approx 3,470.35 , \text{mol} \]
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Using the stoichiometric ratio, calculate the moles of KClO₃ required: \[ \text{Moles of KClO}_3 = \left( \frac{2 , \text{moles KClO}_3}{3 , \text{moles O}_2} \right) \times 3,470.35 , \text{mol} \approx 2,313.57 , \text{mol} \]
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Now, convert moles of KClO₃ to grams: \[ \text{Mass of KClO}_3 = 2,313.57 , \text{mol} \times 122.55 , \text{g/mol} \approx 283,400.8 , \text{g} \]
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Convert grams of KClO₃ to pounds: \[ \text{Mass of KClO}_3 , (\text{lb}) = \frac{283,400.8 , \text{g}}{453.592 , \text{g/lb}} \approx 624.0 , \text{lb} \]
Therefore, to obtain 245 pounds of oxygen, you would need approximately 624 pounds of KClO₃.