#### LIMITING & EXCESS REAGENTS

The following methods are presented to help you better understand limiting reagents, excess reagents  as well as the fundamentals of the stoichiometry process.  The following method presented is the shortest method and most direct method.  It usually takes only a one-step conversion process to obtain information related to a balanced equation.

PROBLEM.  A solution containing 3.50 g of  Na3PO4  is mixed with a solution containing 6.40 g of  Ba(NO3)2.  (a) Which reactant is the limiting reagent?  (b) How many grams of Ba3 (PO4)2 can be formed? (c) How many grams of the excess reagent remain after the reaction is completed. (d) If 2.30 grams of Ba3 (PO4)2 are actually produced in the lab, what is the percent yield for the product?

SOLUTION.   Any solution will require you to write a balanced equation.  Since amounts for both reactants

( Na3PO4  and Ba(NO3)2 ) were given in the problem, this is a limiting reagent problem.   Remember,  these two  reactants will reactant with each other, according to the ratio amounts in the equation, to produce the product until one of the reactants is used up.  Some of the other reactant will still remain un-reacted.  You will need to know which reactant would be used up (the limiting reagent). This will be the reactant that limits the amount of  product that you can form. This will be the reactant that will produce the least amount of product. When this reactant is used up the reaction will stop and there will be some of the other reactant left over.  Remember, there are only two reactants in this equation; one will be the limiting reactant, the other will be the reactant that will have some of it  in excess.

FIRST METHOD.  YOU WILL NEED THE TOTAL GRAMS USED FOR EACH SUBSTANCE IN THE BALANCED EQUATION.  THE COEFFICIENTS WILL BE USED FOR THE MOL RATIOS.

# Total

grams®     328g                        783g                        602g                       510g

2 Na3PO4  (aq)  +  3 Ba(NO3)2 (aq)  ®  Ba3(PO4)2  (s)  +  6 NaNO3 (aq)

2(164g)                    3(261g)                     (602g)                    6(85g)

a.  You must start with the amounts given in the problem for each of the reactants and use the  stoichiometric amounts (in the box) as your converstion factors to determine the amount of Ba3(PO4)2  that  can be produced for each reactant.  Make sure that you use the TOTAL grams for each respective substance.

3.50 g of Na3PO4      X    602 g of Ba3(PO4)2      =             6.42 grams of  Ba3(PO4)2

1                               328     g of Na3PO4

6.40 g of Ba(NO3)2     X   602 g of   Ba3(PO4)2   =               4.92  grams of  Ba3(PO4)2

1                                                                  783 g of  Ba(NO3)2

The reactant that gives the least amount of  Ba3(PO4)2  is the limiting reagent (in this case  Ba(NO3)2).

4.92 grams of Ba3(PO4)2   is all that can be produced.     Therefore the limiting reagent is Ba(NO3)2.

b.  Since  Ba(NO3)is the limiting reagent all of the 6.40 g will be used to form the 4.92 g  of  Ba3(PO4)2    and the

amount of product that it produces will be the only  amount that can be produced from this problem because no

more Ba(NO3)2  remains.  4.92 grams of Ba3(PO4)2   is all that can be produced.

c. There will be some  Na3POleft over however.  To determine this amount you must determine how much Na3PO4

was used  to help make the 4.92 grams of   Ba3(PO4)2 .    To do this you must use the stoichiometric equation to get

4.92  g of  Ba3(PO4)2           X               328  g of Na3PO4     =       0.545  g of  Na3PO4

602  g of  Ba3(PO4)2

0.545 g of Na3PO4   were used to make the 4.92 g of   Ba3(PO4)2 .        Since you started with 3.50 g of the Na3POat the beginning,   a simple subtraction will determine the amount left over

(3.50g, the original amount of Na3PO  -  0.545g, amount of Na3POreacted)    =  2.96 g of Na3PO  is left over.

d. The theoretical yield of the product is the amount that is theoretically calculated using stoichiometry.  The actual

yield is the amount of product that is actually produced in the lab while doing the experiment.

The percent yield = Actual yield           X   100                       2.30 g of Ba3 (PO4)2          X    100     = 46.7 % Percent Yield

Theoretical yield                                      4.92 g of Ba3(PO4)2