Ethanol, C2H5OH, is considered clean fuel because it burns in oxygen to produce carbon dioxide and water with few trace pollutants. If 325.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol were present at the beginning of the reaction?

277.8 grams of ethanol

Explanation:

C2H5OH (l) + 3O2 (g) ⟶2CO2 (g) + 3H2O (l)

It is important to understand that before we can proceed with solving any problem that has to do with stoichiometry, we must first note down the correct and balanced chemical reaction equation.

Number of moles of water formed = mass of water / molar mass of water

Molar mass of water = 18gmol-1

Number of moles of water = 325g/18 gmol-1 = 18.1 moles of water

From the balanced reaction equation;

1 mole of ethanol yields 3 moles of water

x moles of ethanol yields 18.1 moles of water

Hence;

x = 18.1 * 1 / 3 = 6.03 moles of ethanol

Molar mass of ethanol = 46.07 g/mol

Therefore mass of ethanol originally present = 6.03 moles * 46.07 g/mol = 277.8 grams of ethanol