When 20.0 g C2H6 and 60.0 g of O2 react to form CO2 and H2O, how many grams of water are formed?

1 Answer

• 1. Write the balanced chemical equation for the reaction:

  2C2H6 + 7O2 = 4CO2 + 6H2O

  2. Determine moles C2H6 and O2.

  moles C2H6 = 20.0 grams C2H6 x (1 mole C2H6/30.0692 grams C2H6) = 0.665 moles C2H6

  moles O2 = 60.0 grams O2 x (1 mole O2/32 grams O2) = 1.875 moles O2

  3. Determine limiting reagent. The limiting reagent is the reactant producing the smaller amount of product.

  Is it C2H6?

  Note: 2 moles C2H6 produce 6 moles H2O

  0.665 moles C2H6 x (6 moles H2O/2 moles C2H6) = 1.995 moles H2O

  Is it O2?

  Note: 7 moles O2 produce 6 moles H2O

  1.875 moles O2 x (6 moles H2O/7 moles O2) = 1.607 moles H2O

  Therefore, the LR is O2 since it produces the least amount of product.

  3. Convert moles H2O to grams H2O.

  1.607 mole H2O x (18.02 grams H2O/1 mole H2O) = 28.9 grams H2O

  That’s it!n_n