In the examples we did on Wednesday it was assumed one of the reactants was in excess, so the amount of product formed depended on the amount of the other reactant. When the reaction had gone to completion the amount of starting material was zero.

A problem which is more fun involves specifying the amounts of both reactants without indicating which is in excess. It becomes necessary to determine which of the reactants is in excess. Lets begin with a simple example to show the logic. In the reaction,

2C2H6(g) + 7O2(g) ---> 4CO 2(g) + 6H2O (g)

Exactly 2 mols of C2H6 are required to react completely with 7 mols of O2.

Sample Limiting Reagent Problem #1

Sample Limiting Reagent Problem #2

Sample Limiting Reagent Problem #3

Here are two animations showing the second problem using two different approaches. One as we solved the problem in class the other a slightly different way.

Now how about a TERRIBLE PROBLEM!!!!! Try this one. (Not for the faint at heart.)

A gaseous sample with a mass of 5.00 g containing ethane (C2H6) and propane (C3H8) was combusted in excess oxygen. The mass of carbon dioxide produced was 14.9 g. Determine the mass of ethane and propane in the original sample.