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.
Totalgrams® 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.
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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 X 3 mol of Ba(NO3)2 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)2 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 Na3PO4 X 1 mol of Ba3(PO4)2 X 602 g of Ba3(PO4)2 = 6.42 grams of Ba3(PO4)2
1 164 g of Na3PO4 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)
Mi x Vi = Mf x Vf