BALANCING MORE COMPLEX CHEMICAL EQUATION

by clever academyMay 28, 20173 Comments




A chemical formula shows the number of atoms in a molecule. Chemical symbols and formula are combined together to form chemical equations, which explains a chemical equation.
It has been noticed that some chemical equations cannot be balanced using trial and error method. Therefore simultaneous equation method is used. The procedures are as follows;

  •   Put coefficient a, b, c, d…, in front of each chemical formula, starting from reactants to products.
  • For each element in the chemical equation, form a mathematical relation, i.e. L. H. S is equal to R. H. S,L. H. S. represents the total number of atoms of each type in the reaction, and R. H. S. represents the total number of atoms of each type in the product.
  • Assume the first letter to be equal to unity, then find other letter in terms of the first letter that was assumed to be unity.

                   WORKED EXAMPLE

Question one

→C3H5 (NO3)3         N2 + O2 +CO2 +H2O                          

                           SOLUTION


→aC3H5 (NO3)3        bN2 + CO2 + dCO2 + eH2O

For   C:    3a =d   .................... (1)

         H:    5a =2c …………………. (2)
         N:    3a =2b …………………. (3)
         O:     9a =2c + 2d + e …….. (4)

The equations above are then solved simultaneously (either by substitution or elimination)

Let a=1 (NOTE: when using this method the initial letter is always equal to 1)

Substitute the value of a=1 into equation (1), (2) and (3)


           From equation 1

           3a =d               3 x 1 =d           d =3       


                 From equation 2

                       5a =2e

                      5 x 1 =2e         e =5/2

                  From equation 3

                      3 x 1 =2b

                      3 =2b           b =3/2
               
                  From equation (4)

                      9 x 1=2c + 2 x3 + 5/2

                      9 =2c +6 +5/2        2c = 1/2                                                     
                                                                                         
                                     C =¼

 Therefore we have that;

a= 1,       b= 3/2,        c= ¼,        d= 3,          e= 5/2.


C3H5 (NO3)3 →          3/2N2 + ¼O2 + 3CO2 + 5/2H2O

Multiply through by 4 to get the coefficient in whole number integer;


4C3H5 (NO3)3 →         6N2 + O2 + 12CO2 + 10H2O



    IMPORTANCE OF CHEMICAL EQUATION

  •  It tells us the direction of the reaction and whether the reaction is reversible.
  •  It tells us the individual elements and radicals involved.
  •  It gives a mental picture of the movement of the element and radicals during the reactions.
  •  It tells us the state of matter in which the substance are present, as indicated by the state symbols-(S) for solid, (L) for liquid, (G) for gas, (aq) for aqueous solution which are placed after the formulae of the substances in the equation.



SOME COMPLEX EQUATIONS 
K4Fe(CN)6 + H2SO4 + H2O        →      K2SO4 + (NH4)2SO4 + CO + FeSO4

 Fe(CO)5 + NaOH       →      Na2Fe(CO)4 +Na2CO3 + H2O

 CrI3 +KOH + CL2       →    K2CrO4 + KIO4 +KCL +H2O 

 H3PO4 + (NH4)2M0O4 + HNO3    →    (NH4)3PO4 +  M0O4 + NH4NO3 + H2O


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