Example 1: Balance the equation for the reaction of solid potassium nitrate decomposes
when heated to produce solid potassiumm nitrite and oxygen gas.
KNO3 (s) ===> KNO2 (s) + O2 (g)
Step 1: balance K first
It already is. So far not too bad.
Step 2: balance N next
It already is. Boy this is easy.
Step 3: finally, balance O - not so easy, but not too hard.
You have 3 O atoms on the left and 4 O's on the right. How can you balance
the O's w/o changing the coefficients of 1 in front of the KNO3 and KNO2?
(NOTE: you don't write a 1 when it is the subscript in a formula or the coefficient in
front of a formula in an equation.)
What you need to realize is that the 3 O's in the KNO3 and 2 O's in the KNO2
shouldn't be changed directly since that would throw K and N out of balance. So what you
actually have to ask is "How can I balance the 3 O's on the left starting with 2 O's on the
right (in the KNO2)?". Write it like this:
KNO3 (s) ===> KNO2 (s) + O2 (g)
3 O's 2 O's + "How many O's do I need from O2 to give a total of 3?"
The answer of course is you need only 1 O atom from O2 to add to the 2 O's you
already have to give a total of 3 O's on the right. How can you get 1 O atom from an O2?
If you take 1/2 of an O2 molecule that will give you 1 O atom. You use a coefficient
of 1/2 in front of the O2. This will give you the 1 O that you need to balance the O atoms.
So at this point you would have the following:
KNO3 (s) ===> KNO2 (s) + 1/2 O2 (g)
NOTE: Using a fraction isn't that difficult. You needed ONE O atom. That's what goes in
the numerator. The subscript for the O2 is a TWO. That's what's in the denominator.
If it had been an O3 you would have used a fraction with a 3 in the denominator.
While this may not always be the case, the denominator and subscript for the atom your balancing
using the fraction will be related.
However, remember one of the requirements is to have whole number coefficients (Although
there are times we do see fractions. It depends on what you are trying to use the equation
for, but let's not get into that right now.) To get whole numbers you simply multiply the
whole equation by 2. This will give the following:
2 KNO3 (s) ===> 2 KNO2 (s) + O2 (g)
Don't forget there is a "1" in front of the O2.
Step 4: Check to make sure everything is balanced.
Left Right 2 K 2 K 2 N 2 N 6 O 6 O So everything is balanced. This wasn't too tough, was it?
Example 2: Balance the equation for the reaction of solid calcium carbide with
water to produce calcium hydroxide and acetylene gas.
CaC2 + H2O ===> Ca(OH)2 + C2H2 Answer: CaC2(s) + 2 H2O(l) ===> Ca(OH)2(aq) + C2H2(g)
Example 3: Balance the equation for the combustion of acetylene gas. Complete
combustion (also referred to as just combustion) of a hydrocarbon
(contains C & H) results in CO2 and H2O as the products.
C2H2 + O2 ===> CO2 + H2O
Hint: Start with the C. You may want to use a fraction to help you balance the oxygens and
save the O atoms until last. What fraction do you need in front of the O2? Where
do you think the 5 goes in the fraction (you need 5 O atoms)? What goes in the denominator
(see the subscript of 2)? What fraction would you need if the C2H2 reacted
with O3 (ozone)?
Answer:
2 C2H2(g) + 5 O2(g) ===> 4 CO2(g) + 2 H2O(g)
Example 4: Balance the equation for the reaction of solid calcium oxide with
phosphoric acid to produce solid calcium phosphate and water.
CaO + H3PO4 ===> Ca3(PO4)2 + H2O Hint: Balance the PO4 as a unit since it appears in that form on both sides of the arrow. This is the PO43- polyatomic anion. Answer: 3 CaO(s) + 2 H3PO4(aq) ===> Ca3(PO4)2(s) + 3 H2O(l)
Bonus Example: Balance the equation for the reaction of liquid phosphorus trichloride with
water to produce phosphorus acid and hydrocholric acid.
PCl3 + H2O ===> H3PO3 + HCl Hint: Start with the Cl. Answer: PCl3(l) + 3 H2O(l) ===> H3PO3(aq) + 3 HCl(aq)
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