Limiting Reactant

Limiting Reactant:  reactant that will be consumed (used up) first;  limits the amount of product that can be made (produced)

Excess Reactant:  reactant that will not be completely consumed by reacting with all of the limiting reactant;  there will be some of this reactant left over after the reaction is complete

How is a limiting reactant problem different from a regular stoichiometry problem?  There are amounts (grams, atoms/molecules, moles) given with both of the reactants.

How are the calculations different from regular stoichiometry problems?  Once you determine the limiting reactant, the calculations are exactly the same.

How can I determine which is the limiting reactant?  Find the number of moles of each reactant.  Divide the number of moles by its coefficient from the balanced equation.  The smaller of these two numbers is the limiting reactant.  Use the number and unit associated with the limiting reactant to solve the problem.

EX:  (A) How many grams of Fe2O3 can be produced from the reaction of 20.0 grams of Fe with 20.0 grams of O2?  (B) How many grams of excess reactant remains after reaction is complete?
_4_ Fe  +  _3_ O2
à  _2_ Fe2O3

(A) STEP 1:  determine the limiting reactant

Fe: 20.0 g Fe | 1 mole Fe   = 0.358 moles Fe                   O2: 20.0 g O2 | 1 mole O2 = 0.625 moles O2
| 55.85 g Fe                                                                     | 32 g O2

Fe: 0.358 = 0.0895                                                        O2: 0.625 = 0.208
4                                                                                 3

So, Fe is limiting reactant.  Use the information given for Fe to solve the problem.

0.358 moles Fe = x moles Fe2O3               4x = 0.716

4                          2                      x = 0.179 moles Fe2O3

0.179 moles Fe2O3 | 159.7 g Fe2O3 = 28.6 grams Fe2O3
| 1 mole Fe2O3

(B) STEP 1:  Using limiting reactant information, determine the amount of excess reactant used.
0.358 moles Fe  =   x moles O2                4 x = 1.074
4                        3                         x = 0.2685 moles O2 used
0.2685 moles O2 | 32 g O2   =      8.59 g O2 used
| 1 mole
STEP 2:  Subtract number of grams of excess (O2) used from the original amount given in the problem.

20.0 g – 8.59 g = 11.41 g O2 remain

On your own  if 25.00 grams of aluminum sulfate reacted with 25.00 grams of zinc chloride, how many grams of aluminum chloride could be produced?  How many grams of excess reactant remains after reaction is complete?

___ Al2(SO4)3  +  ___ ZnCl2  à  ___ AlCl3  +  ___ ZnSO4