Stoichiometry: The Art of Balancing Chemical Equations
The idea of a balanced equation was mentioned in
Conservation of Mass.
Here we will discuss how to balance an equation, a process which
involves trial and error. In a college level course the student
is given 20 or so problems homework problems to solve.
One must select a reactant and set its coefficient to one, and then select the correct coefficient for the corresponding product. Usually, this will automatically set the coefficient of the other reactant, and then in turn, another product.
For our example reaction, we will combust ethanol
(C2H6O).
a C2H6O + b O2 --> c CO2
+ d H2O
As a hint, if a reactant with just one type of atom is present
(H2, O2, N2, Cl2,
etc.)
save it for last.
The carbon of ethanol can only go to the CO2 product.
The hydrogen of ethanol can only go the H2O product.
The oxygen of ethanol can go to either CO2 or H2
which could make things complicated. Fortunately, once the constants
a, c, and d are set (which will take care of C and H), the total
amount of oxygen in the products can be calculated and from this,
we can calculate b.
Set 'a' to 1.
Since there are two C's per ethanol, this sets 'c' to 2.
Since there are six H's per ethanol, this sets 'd' to 3.
Note there two hydrogens in water, which
changes the math
The equation is now:
1 C2H6O + b O2 --> 2 CO2
+ 3 H2O
From CO2 there are four oxygens, and from H2O
there are three oxygens, for a total of seven oxygens. In calculating
b, we note that there are two oxygens per molecule, so we set b to
(7/2).
1 C2H6O + (7/2) O2 --> 2 CO2
+ 3 H2O
We can multiply each constant by two to remove the fraction.
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