How many moles of o2 are required for the complete reaction of 54.7 g of c2h4 to form co2 and h2o?

The balanced chemical equation that illustrates this reaction is:

C2H4 + 3O2 - - > 2CO2 + 2H2O

From the periodic table:

mass of carbon = 12 grams

mass of hydrogen = 1 gram

Therefore:

molar mass of C2H4 = 12 (2) + 4 (1) = 24 + 4 = 28 grams

number of moles = mass / molar mass

number of moles of C2H4 = 54.7 / 28 = 1.95 moles

From the balanced equation above:

3 moles of oxygen are required to react with one mole of C2H4, therefore, to know the number of moles required to react with 1.95 moles of C2H4, all you have to do is cross multiplication as follows:

number of oxygen moles = (1.95*3) / 1 = 5.85 moles

5.85 moles. Lookup the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Calculate the molar mass of C2H4 2 * 12.0107 + 4 * 1.00794 = 28.05316 g/mol Calculate how many moles of C2H4 you have 54.7 g / 28.05316 g/mol = 1.949869462 mol Now create the balanced equation for the reaction. The unbalanced equation: C2H4 + O2 = = > CO2 + H2O The balanced equation: C2H4 + 3 O2 = = > 2 CO2 + 2 H2O So each mole of C2H4 requires 3 moles of O2. So multiply the number of moles of C2H4 by 3, giving 1.949869462 mol * 3 = 5.849608386 mol Rounding the result to 3 significant digits gives you 5.85 moles.