A
Correspondence 1,
BALANCING CHEMICAL EQUATIONS DEAREDITOR: The method of balancing equations given by A. W. S. Endslow on page EDUCATION 2453 of the December, 1931, issue, of the JOURNAL OF CHEMICAL has been in use in German high schools for a number of years. This method has been explained on page 218 of the seventeenth edition (1919) of a widely used high-school textbook, "Gmndriss der Chemie," by Dr. Fr. RiidoriT. The following example is given: According to which equation will iodic acid react with sulfur dioxide in the presence of water with the formation of sulfuric acid and precipitation of iodine? The equation is: rHIOi then
+ yS02 + zH30 = xH2S04+ 7.1 x
(1)
+ Zz = 2u
(2) x = ll R\: Zy 2e = 4u (3) (4) v = u l e t ~ = l : t h c n z = 2 ; u = ' / ~y;= V 9 ; u = 1 therefore 2 = 2 ; y = j ; z = 4 ,. u = 5 ; u = 2 hence, the equation k 2HIOs 5S02 4HaO = 5HsS01 21.
+ +
+
+
+
The author gives a number of problems, a few of which are quoted below: 114 According to which reaction will pyrolnsite react with potassium iodide and sulfuric acid in such a way that manganese sulfate (MnSOa), potassium acid sulfate (KHSOn), water, and iodine are formed? 118 According to which reaction does a mixture of coal and potassium nitrate bum to potassium carbonate, carbon dioxide, and nitrogen? 125 When zinc is dissolved in dilute nitric acid, zinc nitrate, water, and ammonium nitrate are formed. By which equation can this reaction be expressed? 126 Find the coefficients of the following equations:
++ ++
+ + + +
+ + + + + +
(a)-%As y H 8 0 1 = eAssOa uHnO vSOn yHNOs Z H ~ O= uH3AsOe uNO (b)-zAs20s ySOz zH?O = uHgCl uH2S04 wHCl (k)-.xHgClr YHZSOI zHIS = uMnSOa vKL~OI W H ~ O 6 . (0)-.zKMnOd
+
This method of balancing equations is based on the system used in solving simultaneous linear equations. I n the first example given illustrating the reaction between iodic acid and sulfur dioxide four simultaneous linear equations can be formed containing five unknown constants. If a definite value is assigned to one of these constants then the value of the other four constants can be found easily. The method will fail completely if the number of unknown constants (total number of substances involved in 358
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CORRESPONDENCE
359
the reaction) exceeds the number of simultaneous linear equations by more than one. This is true in the case of Equation 126, ( 0 ) . xKMnO.4- yHISOl for K for Mn for 0 for H for S
+ zHIS = uMnS01 + zK2S04+ wHnO+ rS x = 2u r = u 4% 4y = 4u 4v 2y 2s = 2w y + z = u + v + r
+ +
+
(1)
(2) (3) (4) (5)
+w
The equation involves seven substances, hence, seven unknown constants; only five simultaneous linear equations can be formed by the application of purely mathematical, non-chemical prinaples as only five elements take part in the reaction. It is impossible to solve this set of equations because five simultaneous linear equations containing seven unknown constants cannot be solved. A trial will make this statement more convincing. Let w = 1 ; then n = 2 and u = 2. By substitution we get (.5)
4+4y=8+ 4+w y+e=w y + z = 2 + l + r
(5)
4y=w+r y + s = w y+z=r+3
(3) (4)
I~C~CP (:+I (4) or
".
4y = w + 4 w=r+3 4y=r+7
In this way a linear equation is found containing hwo unknown constants. However, the above equation can be solved if chemical knowledge is utilized in aiding mathematics. The elementary sulfur given on the right-hand side of the equation is formed solely by the oxidation of hydrogen sulfide. This cannot be foreseen by mathematics alone. Hence: z = r
(6)
Now we have six equations with seven unknowns. If we let then (3) (4) (5)
u = l s=2andu=2 4y = w + 4 y + z = w y+z=a+3
From Equation ( 5 ) it follows that y = 3 By substitution of the value of y we get (3) 12 (4)
+
as
hence 2KMnOl
=
w
+ 4;
hence w = 8
3 2 = 8; hence. z = 5 z = r, we get r = 5
+ 3Hd30a + 5H9S = 2MnS01+ KlS04 + 8HaO + 5S
LONGISLAND UNIVERSITY BROOKLYN, NEWYO=
Very sincerely yours, LEOPOLD SCHEFLAN
