To determine how many grams of potassium oxide (K₂O) are produced from the reaction of potassium (K) and oxygen (O), we can follow these steps:
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Identify the chemical reaction: The reaction between potassium and oxygen can be represented by the following balanced equation:
\[ 4 \text{K} + \text{O}_2 \rightarrow 2 \text{K}_2\text{O} \]
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Convert grams to moles:
- Molar mass of potassium (K) = 39.1 g/mol
- Molar mass of oxygen (O) = 16.0 g/mol
For potassium: \[ \text{Moles of } K = \frac{2 \text{ g}}{39.1 \text{ g/mol}} \approx 0.051 moles , K \]
For oxygen: \[ \text{Moles of } O = \frac{5 \text{ g}}{32.0 \text{ g/mol}} \approx 0.156 moles , O_2 , (\text{since \(O_2 = 2 \times \text{O}\)}) \]
Since we need O in \(O_2\) mole for the reaction, the moles of O₂ is: \[ \text{Moles of } O_2 = \frac{5 \text{ g}}{32.0 \text{ g/mol}} \approx 0.156 moles , O_2 \]
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Determine limiting reactant: From the balanced equation \(4 \text{K} + \text{O}_2 \rightarrow 2 \text{K}_2\text{O}\), we can see the mole ratio:
- 4 moles of K react with 1 mole of O₂.
From the available moles:
- For 0.051 moles of K, the amount of O₂ needed is: \[ \frac{0.051}{4} = 0.01275 , \text{moles of } O_2 \]
Since 0.01275 moles of O₂ is much less than the 0.156 moles available, potassium (K) is the limiting reactant.
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Calculate the amount of K₂O produced: From the reaction stoichiometry, 4 moles of K produce 2 moles of K₂O. Therefore, 0.051 moles of K will produce: \[ \text{Moles of } K_2O = \frac{0.051 \text{ moles K} \times 2 \text{ moles } K_2O}{4 \text{ moles K}} = 0.0255 \text{ moles } K_2O \]
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Convert moles of K₂O to grams: The molar mass of K₂O is calculated as: \[ \text{Molar mass of } K_2O = (2 \times 39.1) + 16.0 = 94.2 \text{ g/mol} \]
Therefore, the mass of K₂O produced is: \[ \text{Mass of } K_2O = 0.0255 \text{ moles} \times 94.2 \text{ g/mol} \approx 2.397 \text{ g} \]
Thus, approximately 2.4 grams of potassium oxide (K₂O) will be produced from the reaction of 2 grams of potassium and 5 grams of oxygen.