Chemistry 2

Limiting Reactants

  1. Limiting Reactants
    1. All of the above problems assume you are using only the precise amounts necessary to perform the reaction. Often you measure one of the reactants precisely and react it with an excess of another substance to make sure all of the first substance is consumed in the reaction.
    2. Consider the following problem. If you start with 12.0 mol of Al and 10.0 mol of O2 which would run out first and how much would you have left over of the other substance?
    3. To solve this problem do two calculations, one in which you assume all of the Al is used and one in which all of the O2 is used.
Assuming all of the Al is used
12.0   x    
4 Al + 3 O2 ---> 2 Al2O3


x = 9.00 mol of O2 needed

Assuming all of the O2 is used
x   10.0    
4 Al + 3 O2 ---> 2 Al2O3


x = 13.3 mol of Al needed

    1. As you can see from either calculation, we have more oxygen than can be used in this reaction. Aluminum is the limiting reactant. 1.00 mol of oxygen (10.0 mol - 9.00 mol) will be left over.
    2. To do this given grams you will have to do the g->mol conversions before the above mole ratio calculations.