More on Chemical Reaction Balancing D. F. Swinehart' University of Oregon, Eugene, OR 97403
G. R. Blakley in a recent article2 on chemical equation balancing makes the statement that "only the matrix method is powerful enough to balance the skeletal chemical equation Hz
+ Ca(CNh + NAIF4 + FeSO4 + MgSiOa + KI + HaPo4 + PbCrOh + BrCI + CFzCIz + SO2 = PbBm + CrCIa + MgCOa + KAI(OH)4+ Fe(SCN)3 + PI3 + NazSiOa + CaF2 + H z 0
Despite such a claim, the balancing of this reaction can be accomplished without using the matrix method. We use a method of writing certain partial reactions, some of which are metathetical and some of which are redox half-reactions. All of these are balanced, and properly combining these partial reactions yields the final balanced reaction. We start arbitrarily with the observation that all the Fe and S are found among the products in Fe(SCNh. Therefore we write 6 e- +-S04z-
+ CN- + 8 H+ = SCN- + 4 Hz0
MgSiOa
We note that 3 SCN- are required to make Fe(SCN)a and one Soh2- to match FeS04. Thus we double the last reaction also Fez+= Feat
+ e-
(3)
Combining eqns. (1-3) yields 13 e-
T o match K+ and I- and cancel Na+, Ca2+, and COa2- we calculate 6 X (eqn. (7)) 3 X (eqn. (9)) 3 X (eqn. (8)) 5 X (eqn. (6)) 2 X (eqn. (10)) to give
+
+
+
+
+ 6 NaAIFa + 6 KI + 3 MgSiOa + 2 H3P04+ 10 FeSOa + 20 SO2 + 15 Ca(CN)z+ 3 CFzClz + 128 H+ = 15 CaFz + 6 KAI(0H)a + 3 MgCOa + 3 NazSiOa+ 2 PI. + 10 Fe(SCN)a + 6 CI- + 55 Hz0 (11)
134 e-
The final partial reactions are, apart from the oxidation of Hz to H f , 2 e-
+ BrCl = Br- + C1-
(13)
+ 6 X (eqn. (12)) + 12 (eqn. (13)) yields
176 e- + 6 NaAlFa + 6 KI + 3 MgSiOa + 2 Hap01 + 10 FeSOl + 20 SO2 + 15Ca(CN)z + 3 CFzClz + 6 PbCrO4 + 12 BrCl + 176H+ = 15 CaFz + 6 KAI(OH)4+ 3 MgCOz + 3 NazSiOa + 2 PI3 + 10 Fe(SCN)a + 6 PbBm + 6 CrCIa + 79 Hz0 (14) All the half-reactions have been reductions (electrons added on left). The only oxidation in the reaction is the oxidation of Hz to H+.
+ FeSOa + 2 SOz+ 3 CN- + 16 HC= Fe(SCN)a + 8 Hz0 (4)
We note that all the CN- comes from Ca(CN)z, and we write Ca(CN)z= Caz+ + 2 CN-
3 x (eqn. (5)) 26 e-
(9)
~ ~ - + ~ H + + ~ I - + H ~ P O I = P I ~ + ~ (10) HZO
(eqn. (11)) ~~-+SO~+CN-+~H+=SCN-+ZHZO
+ C0z2- + 2 NaC = MgC03 + NszSiOa
and
(1)
and
(6)
88 Hz + 12 BrCl + 6 PbCr01+ 6 NaAIFd + 6 KI 3 MgSiOa + 2 Hap01 + 10 FeSOa 20 SOz + 15Ca(CN)z + 3 CFzCIz = 15 CaFz+ 6 KAI(OH)I+ 3 MgC08 + 3 NazSiOa + 2 PI3 + 10 Fe(SCNh + 6 PbBh + 6 CrCIa + 79 Hz0
+
Next we observe that all the calcium appears among the products as CaFz. Thus we write
+
2 CaZ+ NaAIFa
+ 4 H20 + Kt = 2 CaFz
+ KAl(0H)a+ Na+ + 4 Hf
(7)
+ C0aZ- + 2 CI- + 6 H+
(8)
and Caz+ + CFzClp+ 3 Hz0 = CaFz
+
88 HP= 176 H+ 176 eand add this to eqn. (14) to give the final balanced reaction.
+ 2 FeSOa + 4 SO2 + 3 Ca(CNh + 32 H+
+ 16 Hs0 + 3 Ca2+
There are 176 electrons and 176 H+ on the left side of eqn. (14). Thus we multiply eqn. (15) by 88 to give
(5)
+ 2 X (eqn. (4)) yields = 2 Fe(SCN)a
'
and
Retired. hesent address: 2312 A Berwick Blvd.. Columbus, OH 43209.
+
As final comments, the time i t took originallyto accomplish the above balancing was a couple of hours. I t probably took Blakeley only a matter of seconds on a computer. However, that does not count the time required to write the program and to trouble-shoot it. The present author used a hand-held calculator to do the simple arithmetic. EOVC., 59, 728 (1982). Blakeley, G. R., J. CHEM.
Volume 62 Number 1 January 1985
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