Gaseous ethane (CH3CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O)

Question

Suppose 27.4 g of ethane is mixed with 42. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Whenever you are given the amounts of both reactants, the first thing (after balancing the equation) is to find which reactant is limiting.

CH₃CH₃ + 3O₂ ==> 2CO₂ + 2H₂O ... balanced equation

To find the limiting reactant, one way is to simply divide moles of each by the corresponding coefficient in the balanced equation.

For CH₃CH₃ we have 27.4 g x 1 mol / 30 g = 0.913 mol CH₃CH₃ (÷1->0.9)

For O₂ we have 42. g O₂ x 1 mol / 32 g = 1.31 mol O₂ (÷3->0.4)

Since 0.4 is less than 0.9, O₂ is the limiting reactant, and we use the moles of O₂ to find the theoretical yield

Theoretical yield of H2O = 1.31 mol O₂ x 2 mol H2O / 3 mol O₂ x 18 g H2O/mol = 15.7g H2O

Theoretical yield of H2O = 16. g H2O (2 sig. figs. based on the value of 42. g O2)