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.
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)2 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 Na3PO4 left 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 Na3PO4 at the beginning, a simple subtraction will determine the amount left over
(3.50g, the original amount of Na3PO4 - 0.545g, amount of Na3PO4 reacted) = 2.96 g of Na3PO4 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