Reaction Stoichiometry and Suitable "Coordinate Systems" R. J. Tykodi Southeastern Massachusetts University, North Dartmouth, MA 02747
Our physicist colleagues, when dealing with (nonrelativistic) dynamics, give their students a secure rock to stand on (force = mass X acceleration) and urge upon them a flexible approach to the use of coordinate systems-Cartesian coordinates are good here, spherical polar coordinates are good there, and so on. A flexible approach to the use of prohlem-solving techniques is the mark of a good problem-solver. Too many of our general chemistry students get locked into one rigid format for dealing with reaction stoichiometry and go astray if the prohlem-solving road is the least bit winding.',= I suggest that we follow the lead of our physicist colleagues and, when dealing with reaction stoichiometry, introduce our general chemistry students to several of the standard problem-solving techniques-pointing out t h e special strengths of each one. Let us stress flexibility in the use of prohlem-solving tools. Consider the following three methods for dealing with problems involving reaction stoichiometry.
4FeS,
+ 110,
(2) What is the maximum number of moles of SO2 that can he produced from 7.0 rnol of FeS2?
Divide the reaction equation through by 4 to see what the mole ratios are for the consumption of 1mol of FeSz (unitize): 11 FeSz + -0, 4
-
2 - Fe,O,
4
+ 8 SO,
and then multiply the resulting equation through by 7.0 to scale up to the required 7.0 rnol of FeS2:
Wr see immedinrclp that upon rhr coniumption of: 11mol of FeS, I ~ maxilnum P number of mole~olSO. that mn br pndnred is 7.0 X
Unitize and Scale Up'33
For the reaction
We see immediately that to produce 5.0 mol of Fe%Oa requires 5.0 X (1112)mol of 02.
-
(8/4). 2Fe,03 + 880,
(1)How many moles of 0 2 are needed to produce 5.0 mol of FenOa,given an excess of FeSz?
Divide the reaction equation through by 2 to see what the mole ratios are for the production of 1 mol of Fez03 (i.e., unitize): 4 11 B - FeS, + -0, Fe,O, +-SO, 2 2 2
-
and then multiply the resulting equation through by 5.0 to scale up to the required 5.0 mol of FelOs:
eood The unitize and scale U D technioue is an esneciallv " " one for establishing mole ratios among the reactants consumed and the products formed in agivenchemical reaction. Factor-Label Procedure
Consider the reaction
How many grams of oxygen do we need t o produce 50.0 g of
NO=?
' Beichel, G. J . J. Chem. Educ. 1986, 63,146.
Herron, J. D.; Greenbowe, T. J . J. Chem. Educ. 1986, 63,528. Navidi. M. H.; Baker. A. D. J. Chem. Educ. 1984,61.522.
958
Journal of Chemical Education
Let 5 stand far the concept of reaction-equiualence, that is, connects quantities that are equivalent in terms of the given reaction. Then,
1mol NO, 1mol0, 50.0 g NO, = 50.0 g NO, (46.01 NOz) (2 rn0l NO)
0, (m) 32.00 g
that is, 50.0 g of NO2 is readion-equivalent to 17.4 g of Oz in the indicated reaction.
(1)For the reaction,
how many grams of oxygen do we need t o produce 50.0 g of NO?? This is the problem we treated before via the factorlabel method. The de Dander relation tells us that
In the factor-label procedure we operate on the input information with a string of factors containing appropriate labels in such a way as to cause cancellation of all unwanted labels-yielding in the end the sought-after quantity. The factor-label procedure is especially good for converting from one unit of measure to another-from inches to centimeters, from angstroms to nanometers, etc.; it is also good for mass-mass problems and for problems involving percent composition. De Donder Ratlos Suppose we write a general balanced chemical reaction equation this way:
where what we mean is that a1 mol of substance A1 reacts with a, mol of substance A2 and with . . . to form PI mol of substance B, and B2 mol of substance Bg a n d . . . . Let n'(. . .) stand for the number of moles of a reactant consumed, or the number of moles of a product formed, in the reaction we happen to be studying. The fundamental d e Donder relation is that
and
(2) Consider the reaction,
We add 35.0 g of BaCl, and then 15.0 g of H2S04 to a large amount of water (say 2 kg); how much BaS04(s) precipitates from the solution (at room temperature)? The de Donder relation tells us that n'(BaC1,) - n'(H,SO,) -1
I call the quantities n'(AJ/ai, n'(Bj)lp, d e Donder ratios; the de Donder relation says that all the de Donder ratios for the reactants and the products in a given chemical reaction are equal to one another. The symmetry of the de Donder relation makes it easy t o remember. For the reaction
the de Donder relation says that nf(FeS,) - n'(O3) 4 11
- n'(Fe,OJ ---n'(S0,) 2
8
-
1
n'(BaS0,) 1
In our earlier examples we assumed, in effect, that we had available as much as we needed of the substances not specifically mentioned by mass; thatis, weassumed thatwe had availableexactlythe right amount or a surplus of each of the other substances. In this example we have definite masses of both BaClz and H2SOa.Which one is the controlling reagent-the one that establishes the size of the de Donder ratios for the reaetion-and which one is present in excess? The procedure to follow is this: calculate apparent reaction ratios [ # mol availablelcoeff.] from the given masses; the smallest of the apparent reaction ratios [# mol availablelcoeff.] is the true de Donder ratio In'(. . .)lcaeff.];use the true de Donder ratio to solve the problem. Following this procedure we get
# rnol of BaCI,
and for the reaction
1
# mol of H,SO, the de Donder relation says that
1
=
35.0 FM(BaC1,)
=35'0 = 0.168
15.0 FM(H,SO,)
- 15.0 - 0.153 -
208.2
The apparent reaction ratio for HzSOI is the smaller one, so it is the true de Donder ratio; therefore, We are indebted to the Beleian scientist T h e o ~ h i l ede I)onder (187:1-1957) for pointing out to us the elegance nnd utilitv ot the reartion-ratio method for dealing with stoichiometric relations. 1 ncrepf with graritude dtt Vonder's basic idea.' bur I do not use his ferminolorv or his s m b d i s m (for another "Americanization" of de ~ o K d e r ' sidea, see the article by Garsts). Let us apply the reaction-ratio method to a few examples. Let F M ( . . .) represent the formula mass (molecular weight) of the suhstance shown in the parentheses. ~
~
De Donder. Th. Legons de Thermodynarnique et de CherniePhysique;Gauthier-Villars: Paris. 1920. Garst. J. F. J. Chem. Educ. 1974, 51. 194.
and
# g of BaSO, FM(BaS0,)
= 0.153
SO
# g of BaSO,
= n'(BaS0,) X
FM(BaSOI) = 0.153 X 233.4 = 35.7
The BaCIz was in excess; only so much of the BaClz reacted as to make n'(BaClz)/l = 0.153; there was, therefore, 0.168 - 0.153 = 0.015 ma1 of BaClz left over (i.e., unreaeted). Note how smoothly the de Donder ratio method handles limiting-reagent problems. Volume 64
Number 11 November 1987
959
Tltrailon Relations
When we titrate a sample of acid or base with a standardized solution of base or acid, the titration reaction (in aqueous solution) is a Acid
+ B Base
-
o Salt
and
+ w Water
can be zero), and the de Donder relation for the reaction tells us that n'(Acid) - n'(Base) (w
a
P
If we measure both the amounts of the acid and the base via calibrated glassware (buret, pipet, etc.), the de Donder relation takes a symmetric form: Molarity(Acid)X # IiterdAcid)
lo3
x -= 0.0325 mol L-' 38.41 Proceeding uia Partial Steps
# males of HCI used = 0.108 mol L-'
C(
# moles of Ba(OH), used = 0.002495 mol HCI or, equivalently, Molarity(Acid)X Volume(Acid)
1mol Ba(OH)%
2 mol HCL
n
= 0.001248
This number of moles of Ba(0H)s is distributed through 0.03841 L, so with the volumes expressed in any convenient (but common to both sides) unit. If we start with a solid sample of acid or base, dissolve it in water, and titrate the resulting solution with a standardized base or acid, the de Donder relation n'(Acid)la = n'(Base)lB has Molarity X # literslcoeff. on one side and # grams/(formula mass X coeff.) on the otber-a perfectly good, but nnsymmetric, form of the d e Donder relation. Example
M(Ba(OH),) =
# moles Ba(OH)% =--0'001248 # liters Ba(OH), 0.03841
- 0.0325 mol L-'
Of the methods we have used to solve this problem, the de Donder ratio method is especially clear and efficient. Also, note that, if we have a balanced reaction equation to work from, acid-base titration calculations can be carried through in a simple and straightforward fashion in terms of molarity-it is not necessary to use the normality concept; the same is true of redox titration calculations. Summary
It takes 23.10 mL of 0.108 M HCI to titrate 38.41 mL of Ba(OH)z solution. What is the molarity of the Ba(OH)?solution? This is an example of the "symmetric case" discussed above: 0.108 x 23.10 - Molarity(Ba(OH),) X 38.41 2 1
Calculations involving the "unsymmetric case" are just as easy. Let us also carry through a factor-label analysis of the problem-first formally, concentrating on cancelling unwanted labels, and then less formally via partial steps. Formal Procedure We may say that 23.10 mL(HC1) = 23.10 mL(HC1)
X
960
1 mol Ba(OH)% 2 mol HCI
Journal of Chemical Education
mol HCI
Just as in certain problems in physics the matching of the symmetry of a coordinate system to the inherent symmetry of the problem makes for ease in solving the problem, so let us stress for our general chemistry students flexibility in attacking prohlems related to reaction stoichiometry. Let us expose our students to several methods of dealing with reaction stoichiometry and let us teachgood judgment in matching method to problem: (1) unitize and scale up-a good and efficient method for estahlishing mole ratios among the reactants consumed and the products formed in a given chemical reaction; (2) factor-label procedure-an excellent method for converting from one unit of measure to another, and a good method for handling mass-mass problems and percent composition prohlems; (3) de Donder ratios-a secure "rock" on which to stand when attacking any problem involving reaction stoichiometry, but especially fine for dealingwith titration relations and with "cantrolling reagent" problems. A final word of caution: let us not "oversell" the factorlabel idea. Students can get the idea that the factor-label procedure is a panacea that will mechanically solve all their problems, and they can thereupon (unthinkingly) apply factor-label procedures in inappropriate
