To determine how many grams of potassium oxide (K₂O) will be produced from the reaction of 2 grams of potassium (K) and 5 grams of oxygen (O), we first need to look at the chemical reaction that occurs between potassium and oxygen:
\[ 4K + O_2 \rightarrow 2K_2O \]
According to the balanced chemical equation, 4 moles of potassium react with 1 mole of oxygen to produce 2 moles of potassium oxide.
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Convert grams to moles:
- Molar mass of K (potassium) = 39.10 g/mol
- Molar mass of O (oxygen) = 32.00 g/mol
- Molar mass of K₂O = 39.10 * 2 + 16.00 = 94.20 g/mol
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Convert grams of K and O to moles:
- Moles of K from 2 grams: \[ \text{Moles of K} = \frac{2 \text{ g}}{39.10 \text{ g/mol}} \approx 0.051 moles \]
- Moles of O from 5 grams: \[ \text{Moles of O} = \frac{5 \text{ g}}{32.00 \text{ g/mol}} \approx 0.156 moles \]
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Determine the limiting reagent:
- From the balanced equation, 4 moles of K react with 1 mole of O.
- For 0.051 moles of K, the moles of O required: \[ \text{Required O} = \frac{0.051 \text{ moles K}}{4} \approx 0.01275 \text{ moles O} \]
- Since we have 0.156 moles of O available, K is the limiting reagent.
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Calculate the amount of K₂O produced:
- According to the stoichiometry of the reaction, 4 moles of K produce 2 moles of K₂O.
- Therefore, \[ \text{Moles of K₂O} = \frac{0.051 \text{ moles K}}{4} \times 2 \approx 0.0255 \text{ moles K₂O} \]
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Convert moles of K₂O to grams: \[ \text{Mass of K₂O} = 0.0255 \text{ moles} \times 94.20 \text{ g/mol} \approx 2.40 \text{ g} \]
Given that the Law of Conservation of Mass states that the mass of the reactants equals the mass of the products and from our calculations, the total mass of the products would be the mass of K + O, which is 2 g + 5 g = 7 g.
Thus, the answer is 7 grams of potassium oxide (K₂O) will be produced.
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