2.89 moles of water would be produced by the complete combustion of 44.5 grams of ethanol in the presence of excess oxygen.
- According to this question, the following reaction between ethanol and oxygen is given:
- C₂H₆O(l) + O₂(g) → CO₂(g) + H₂O(l)
- However, this equation is not balanced. The balanced equation is as follows:
- C₂H₆O + 3O₂ → 2CO₂ + 3H₂O
- According to this equation, 1 mole of ethanol produces 3 moles of water.
- We convert 44.5g of ethanol to moles by dividing by its molar mass as follows:
- mole = 44.5g ÷ 46.07g/mol
- mole = 0.966mol of ethanol
- If 1 mole of ethanol produces 3 moles of water.
- 0.966mol of ethanol will produce (0.966 × 3) = 2.89mol
- Therefore, 2.89 moles of water would be produced by the complete combustion of 44.5 grams of ethanol in the presence of excess oxygen.
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