Respuesta :
Answer:
To find the limiting reactant and the mass of potassium oxide produced, we need to determine which reactant will be consumed first.
First, we need to find the molar masses of potassium (K) and oxygen (O₂):
- The molar mass of potassium (K) is approximately 39.10 g/mol.
- The molar mass of oxygen (O₂) is approximately 32.00 g/mol.
Next, we'll calculate the number of moles of each reactant:
- Moles of potassium (K) = mass of potassium / molar mass of potassium = 37 g / 39.10 g/mol ≈ 0.946 moles
- Moles of oxygen (O₂) = mass of oxygen / molar mass of oxygen = 123 g / 32.00 g/mol ≈ 3.84 moles
Now, we'll use the stoichiometry of the reaction to see which reactant is limiting:
2K + O₂ → 2KO₂
From the balanced equation, we see that 1 mole of oxygen (O₂) reacts with 2 moles of potassium (K). So, for 3.84 moles of oxygen, potassium would need 3.84/2 = 1.92 moles. Since potassium has fewer moles (0.946 moles), it's the limiting reactant.
Now, we'll calculate the mass of potassium oxide (KO₂) produced using the limiting reactant:
- Moles of KO₂ produced = moles of limiting reactant (potassium) = 0.946 moles
- Mass of KO₂ produced = moles of KO₂ produced × molar mass of KO₂
≈ 0.946 moles × (39.10 g/mol + 32.00 g/mol)
≈ 0.946 moles × 71.10 g/mol
≈ 67.28 g
So, approximately 67.28 grams of potassium oxide (KO₂) will be produced.
Explanation:
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