#### TWO METHODS OF STOICHIOMETRY & LIMITING REAGENT

Two methods of stoichiometry are presented for your perusal.  The FIRST METHOD is the method presented in class.  It is the shortest method and most direct method; It usually takes only one step.  The SECOND METHOD is the traditional method taught.  You should be familiar with both so that you will NOT get them confused.

PROBLEM.  A solution containing 3.50 g of  Na3PO4  is mixed with a solution containing 6.40 g of  Ba(NO3)2.  How many grams of Ba3(PO4)2  can be formed?

SOLUTION.   Both solution methods will require you to write a balanced equation. Amounts  for both reactants

( Na3PO4  and Ba(NO3)2 ) were given in the problem, therefore you will need to know which reactant was used up first (the limiting reagent). This will be the reactant that limits the amount of  product that you can form.

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                       574g

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

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

1.  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.  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

This problem is essentially finished with these two steps.  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.

SECOND METHOD. YOU WILL NEED  ONLY THE GRAMS FOR 1 MOL OF EACH SUBSTANCE IN THE EQUATION.  THE COEFFICIENTS WILL BE USED FOR THE MOL RATIOS.

 2 Na3PO4  (aq)  +  3 Ba(NO3)2 (aq)  ®  Ba3(PO4)2  (s)  +  6 NaNO3 (aq)          (164g)                    (261g)                         (602g)                  (85g)

In this method you must determine the limiting reagent. Choose the amount given in the problem for either reactant and determine the stoichiometric amount  needed for the other reactant to react with that amount.

We could determine how much  Na3PO4   is needed to react with 6.40 g of  Ba(NO3)2   or

We could determine how much Ba(NO3)2 is needed to react with 3.50 g of  Na3PO4

3.50 g of Na3PO4      X    1 mol of    Na3PO4    3 mol of Ba(NO3)X      261 g of Ba(NO3)2     =   8.36 grams of  Ba(NO3)2

1                       164 g of Na3PO4              2 mol of    Na3PO4                 1 mol of Ba(NO3)2

.

We have determined that 3.50 g of Na3PO4  stoichiometrically requires 8.36 grams of  Ba(NO3)2 for a complete reaction.   Since the problem gives only 6.40 grams of  Ba(NO3)2.  Ba(NO3)is therefore the limiting reagent. You must then determine the amount that the 6.40 grams of  Ba(NO3)2    will produce according to the equation.

6.40 g of Ba(NO3)2   X  1 mol of Ba(NO3)2   X  1 mol of Ba3(PO4)2   X  602 g of   Ba3(PO4)2   =   4.92  grams of  Ba3(PO4)2

1                      261 g of  Ba(NO3)2      3  mol of  Ba(NO3)2      1 mol of   Ba3(PO4)2

4.92 grams of Ba3(PO4)2   is all that can be produced.

ALTERNATE METHOD FOR DETERMINING THE LIMITING REAGENT .  Determine the amount of    Ba3(PO4)2      produced for each of the two reactants, using the 2nd  Method.  The reactant producing the least amount of the product will be the limiting reagent and this amount will be the solution to our problem.

3.50 g of Na3PO4      X    1 mol of    Na3PO  X   1 mol of   Ba3(PO4)   X    602 g of Ba3(PO4)2   =   6.42 grams of  Ba3(PO4)2

1                       164 g of Na3PO        2   mol of  Na3PO4                        1 mol of   Ba3(PO4)2

6.40 g of Ba(NO3)2   X  1 mol of Ba(NO3)2   X  1 mol of Ba3(PO4)2   X  602 g of   Ba3(PO4)2   =   4.92  grams of  Ba3(PO4)2

1                      261 g of  Ba(NO3)2      3  mol of  Ba(NO3)2      1 mol of   Ba3(PO4)2

## IN CHAPTER 4,  WE INTRODUCED TWO OTHER EQUATIONS TO SOLVE PROBLEMS WHEN THE REACTANT IS IN A SOLUTION.  We already know that

Mols (n) =   .   g  .

MW

n = M  X  L  (n = mols ; M = Molarity;  ;  L = liters)

Mx  Vi    =  Mx  Vf