Balancing HARD TO Balance Equations PDF

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A comprehensive reviewer on balancing hard to balance equations...

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BALANCING HARD TO BALANCE EQUATIONS COMBUSTION REACTIONS: Many combustion reactions are hard to balance because of the oxygen in both the CO 2 and H2O. Often a mixed fraction of ½ for the O 2 will balance the equation. Only thing is you can’t have a fra ction for a coefficient. Therefore, you double ALL the coefficients when this happens. Example: Balance C 4H10 + O 2 → CO2 + H2O  After doing the C and H you end up with the following. C4H10 + O2 → 4CO2 + 5H2O 

Since there are 13 oxygens on the product side, the coefficient for O 2 could be 6½. C4H10 + 6½O2 → 4CO2 + 5H2O



Now double all the coefficients and the equation is done. 2C4H10 + 13O2 → 8CO2 + 10H2O

PROPORTIONS METHOD: This method should only be done when the equation is too hard to do by inspection. Always start out doing the equation by inspection first. If after several minutes of not getting anywhere, use the proportions method. Example #1: Balance Cu + HNO3 → Cu(NO 3)2 + NO + H2O STEPS: 1) Place variables in front of each substance to represent the coefficients. a Cu + b HNO3 → c Cu(NO 3)2 + d NO + e H2O 2) Since the Law of Conservation of Matter tells us that the same number of atoms of each element must be found on both sides of the equation, write an equation, using the coefficient variables and subscripts, for each element so that the number of reactant atoms are equal to the number of product atoms. Cu H N O

a=c b = 2e b = 2c + d 3b = 6c + d + e

3) Take the longest equation, and by substitution get the equation to be a proportion containing only two different variables. This is the step that will differ from one person to another simply because of the choice of variables you use —as long as you can get two variables by using correct algebra it will work. Since b = 2e, Then: 2e = 2c + d 2e − 2c = d Then:

3(2e) = 6c + (2e − 2c) + e 6e = 4c + 3e 3e = 4c e 4  c 3

4) From this proportion, e = 4 and c = 3. Now you can find the other variables using the equations you found for each element. a=c → a=3 b = 2e → b = 2(4) = 8 d = 2e − 2c → d = 2(4) − 2(3) = 2 5) Now place these values in the chemical equation for the coefficient variables and you are done. 3Cu + 8HNO3 → 3Cu(NO 3)2 + 2NO + 4H2O

Example #2: Balance HIO3 + FeI2 + HCl → FeCl3 + ICl + H2O a HIO3 + b FeI2 + c HCl → d FeCl3 + e ICl + f H2O H I O Fe Cl

a + c = 2f a + 2b = e 3a = f b=d c = 3d + e

a + c = 2f a + 2b = e

→ →

a + c = 2(3a) a + 2d = e

c = 3d + e

→

5a = 3d + (a + 2d) 5a = 5d + a 4a = 5d a 5  d 4

b=d c = 5a

→ →

c = 5(5)

→

e = a + 2d f = 3a

→ →

e = 5 + 2(4) f = 3(5)

→ →

→

c = 5a

a=5 b=4 c = 25 d=4 e = 13 f = 15

5HIO3 + 4FeI2 + 25HCl → 4FeCl3 + 13ICl + 15H2O...