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?

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

1.solid

2.carbon dioxide

3.helium

the answer is 1 i believe