The balanced reaction: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
Solution:
We first convert volume of C2H6 to no. of moles. We use the conditions at STP where 1 mol = 22.4 L thus,
Moles C2H6 = 16.4 L/ 22.4 L =0.7321 mol
In order to determine the limiting reagent, we look at the given amounts of the reactants.
0.7321 mol C2H6 (7/2 mol O2 / 1 mol C2H6) = 2.562 mol O2
From the given amounts of the reactants, we can say that O2 is the limiting reactant since we need 2.562 mol O2 to completely react the given amount of C2H6. The excess reagent is C2H6
To calculate for the amount of products and excess reactants:
0.980 mol O2 (2 mol CO2 / (7/2 mol O2)) = 0.56 mol CO2 (22.4 L / 1 mol ) = 12.544 L CO2
0.980 mol O2 (1 mol C2H6 / (7/2 mol O2)) = 0.28 mol C2H6
Excess C2H6 = 0.7321 mol - 0.28 mol C2H6 = 0.4521 mol C2H6
We then use the molecular weight of C2H6 to convert the excess amount to grams.
0.4521 mol C2H6 (30.08 g C2H6 / 1 mol C2H6) = 13.60 g C2H6
Since the limiting reagent is O2 there will be no oxygen atoms that will be left after the reaction.
