Final answer:
To react with 7.50g of ethanol, approximately 15.6 grams of O₂ are needed.
Step-by-step explanation:
To answer your question, we need to first determine the balanced chemical equation for the reaction between ethanol (C₂H₅OH) and oxygen (O₂). The balanced equation is:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
From the balanced equation, we can see that for every 1 mole of ethanol, we need 3 moles of oxygen gas. To calculate the grams of O₂ required to react with 7.50g of ethanol, we can use the molar mass of O₂ (32 g/mol) and the molar mass of ethanol (46 g/mol).
First, calculate the moles of ethanol:
(7.50g ethanol) / (46 g/mol ethanol) = 0.163 moles ethanol
Next, use the mole ratio from the balanced equation to calculate the moles of O₂:
0.163 moles ethanol * (3 moles O₂ / 1 mole ethanol) = 0.488 moles O₂
Finally, calculate the grams of O₂:
0.488 moles O₂ * (32 g/mol O₂) = 15.6g O₂
Therefore, you would need approximately 15.6 grams of O₂ to react with 7.50g of ethanol.