Answer:
80.9 g of ethanol were present at the begining of the reaction
Explanation:
In a combustion reaction you should know that the formed products are water and carbon dioxide. One of the reactants is always oxygen.
The combustion reaction for the ethanol has this equation:
C₂H₅OH (l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)
We determine the moles of produced water
95 g / 18 g/mol = 5.28 moles . Now we can propose this rule of three:
3 moles of water are produced by 1 mol of ethanol
Therefore 5.28 moles of water were produced by (5.28 . 1) /3 = 1.76 moles of ethanol
Finally we should convert the moles to mass in order to find out the grams that were present at the beginning of the reaction
1.76 mol . 46 g /1 mol = 80.9 g
