Ethanol, C2H5OH, is considered a clean fuel because it burns in oxygen to produce carbon dioxide and water with few trace pollutants. If 95.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol were present at the beginning of the reaction?

I got 80.9g for this question

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