A Simpler Method of Chemical Reaction Balancing W. Harjadi lnstiiut Pertanian Bogor, Bogor, Indonesia Two recent articles in THIS JOURNAL^,^ present different solutions t o the problem of balancing the rather awesomelooking skeletal chemical equation H,
+ Ca(CN), + NaAIF4+ FeS04+ MgSiO, + KI + H,PO, + PbCrO, + BrCI + CF,C12 + SO, = PbBr, + CrCl, + MgCO, + KAI(OH), + Fe(SCN), + PI, + Na2Si03+ CsF, + H,O
We start with P I ( 6 7 and assign the coefficient 1to it; we obtain 3 for KI (61.1 for H~P04(7),3 for KAKOH) (4*), and so on as shown in the sequence below. In actual practice i t is best to enter the coefficients underneath the respective suhstances; this enables us to keep track of the progress and to see which step to take next.
Blakleyl uses the matrix method and Swinehart2 breaks up the equation into groups of partial reactions, balancing and recombining them to yield the final product. Swinehart attempts to show that the matrix method is not the only one powerful enough to balance the reaction. The following is yet another approach that has distinct advantages: it does not take as much time as Swinehart's (a couple of hours according to his article), and it is much easier to teach and quite easy to understand even by the beginner. I t is essentially an extension of the familiar method of "balancing by inspection" and one that I dub the "pingpong method". Balancing by inspection hinges on the fact that certain elements in the reaction appear only in one reactant and one product each; the example below indicates, using italics, the elements in question KI + KIO,
+ HCl = I, + H,O + KC1
Balancingfor oxygen, the coefficients 1and 3 are assigned to KI03 and Hz0, respectively. Subsequently, based on hydrogen and then on chlorine, the value of 6 can he given HC1 and KC1. T o find the coefficient for KI we refer to both KI03 and KI since K is found in both these substances; then, based on the coefficients of KI03 and KI the coefficient for 1 2 is determined. The sequence of assigning these coefficients is lKIO.-3H,O-6HC1-6KCll
Here the elements in italics are the ones used to determine the coefficient of the substance next in the series. Let us now turn to Blakley's skeletal equation. For ease of locating substances we number them (numbers marked with asterisks are for products). We also italicize the elements that appear in only one reactant and one product. Note that SiOs in (5) and (7") and CN in (2) and (5*)can each he treated as a single unit, enabling us to assume that the C in (:;*) has its origin in (10). (3) H, + Ca(CM2+ NaAIF, (1)
(2)
(I*) (2') = PbBr, + CrCL,
(5) + FeSO, + MgSiO, (4)
(3*)
(6) + KZ
(4*)
(5*)
(69
Substances (4) and (5*)are on opposite sides of the equation so the coefficient for SO? is derived from the difference I,rtwt!m the n u m h r r o f ~ a t o &in (4)and (5'). (Seealsostep 16and sreps :!and 12 in the second attempt at halanring.this reaction.). With step 12 the chain is broken and we have to start a new one. The relationship between the values of the coefficients in the sequence to those already assigned being unknown, we must use indefinite values for the new series. Let us start with PbCrOa (8) and assign to i t the coefficient a. We obtain
We have already assigned the coefficient 1%to CF2C12 (st. 7), so a = 3. Now only Hz and Hz0 remain unbalanced. Assigning b to Hz and c to H20,we find from HZ(I), 1H3P04 (7) (st. I), 3 KAI(OH)I ( 4 * ) (st. 2) and H 2 0 (9*)that c = b 4% by balancing the hydrogen atoms. Then, balancing the oxygenatoms in (41, (7). (8),(111, (3*), (4'1, and (9*) yields b = 44 and c = 39%. Multiplying all by 2 results in the balanced product
+ MgC03+ KAlfOH), + Fe(SCN), + PI3
' Blakley, G. R. J. Chem. Educ. 1982, 59,728.
Swinehart, D. F . J. Chem. Educ. 1985, 62,55.
978
Journal of Chemical Education
+ 2 H3P0, + 6 PbCrO, + 12 BrCl + 3 CF2C12+ 20 SO, + 2 PI, + 3 Na2Si03+ 15 CaF, + 79 H,O Time required to balance the reaction: slightly less than 20 min, without the aid of even a hand-held calculator. In principle we can start with any substance other than the one we used. With luck we might even have an easier time. We want asuhstance with as large an indexas possible. Let us see what happens when we start with PhBrz (I*).
Only Hz (1) and H20 (9*)remain without coefficient, achieved without recourse to a second sequence. Because of the experience gained from working the first attempt, this one took less than 10 min.
Volume 83 Number 11 November 1986
979
