If 2.32 g of ethanol reacts with 10.6 g of oxygen, how many moles of water are produced?

0.151 moles Ethanol is C2H6O, and the balanced reaction is C2H6O + 3O2 = = > 2CO2 + 3H2O Now calculate the molar mass of C2H6O. Start with the atomic weights. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass C2H6O = 2 * 12.0107 + 6 * 1.00794 + 15.999 = 46.06804 g/mol Molar mass O2 = 2 * 15.999 = 31.998 g/mol Determine how many moles of each reactant is available. Moles C2H6O = 2.32 g / 46.06804 g/mol = 0.050360293 mol Moles O2 = 10.6 g / 31.998 g/mol = 0.331270704 mol Looking at the balanced equation and the number of moles of each reactant, it's obvious that the limiting reactant is C2H6O, and since each mole of C2H6O will produce 3 moles of H2O, then we just need to multiply the number of moles of C2H6O we have by 3, giving 0.050360293 mol * 3 = 0.151080879 moles And of course, round to 3 significant figures, giving 0.151 moles.