
CHEMICAL REACTIONS
8.5  Calculations based on Chemical Equations
In section 8.2, we learned
that in balancing chemical equations, coefficients can be inserted into the chemical equations so that
the number of atoms of each type on the LEFTHANDSIDE 
= 
the number of atoms of each type on the RIGHTHANDSIDE 
With these coefficients, we can write a ratio of any two or more substances in the chemical equation.
Let's take a look at the balanced chemical equation for the synthesis of water:
2 H _{2} (g) + O _{2} (g) 2 H _{2}0 (l) * 
* The coefficient
in front of O_{2} (g) is by default "1".
Coefficient of "1" is usually not written in.
Here are some ratios that we can write from the
above balanced chemical equation:
 2 molecules of H_{2} (g) requires 1 molecule of O_{2} (g) to produce 2 molecules
of H_{2}0 (l).
The important thing is, the ratio is 2:1:2 for the substances H_{2} : O_{2} : H_{2}0 .

In
any combination of any two substances, the coefficients in the balanced
chemical equation tells us: 
the ratio between H_{2} : O_{2} is always 2:1. 
the ratio between H_{2} : H_{2}0 is always 2:2,
or 1:1. 
the ratio between O_{2} : H_{2}0 is always 1:2. 
 If we scale up 2 times, then:
4 molecules of H_{2} (g) requires 2 molecules
of O_{2} (g) to produce 4 molecules of
H_{2}0 (l).
This ratio of 2:1:2 is
always maintained no matter how much we scale up.
 If we scale up 12 times, then:
2 dozens molecules of H_{2} (g) requires 1 dozens molecules
of O_{2} (g) to produce 2 dozens molecules of H_{2}0 (l).
This ratio of 2:1:2 is
always maintained no matter how much we scale up.
 If we scale up to a mole, then:
2 moles of H_{2} (g) requires 1 mole of O_{2} (g) to produce 2 moles of H_{2}0
(l).
This ratio of 2:1:2 is
always maintained no matter how much we scale up.
I repeat:
The important thing is, the ratio is
always 2:1:2 for the substances H_{2} : O_{2} : H_{2}0 .
The coefficients give the mole ratio of the reactants and products.

In any combination
of any two substances, the coefficients in the balanced chemical equation
tells us: 
the ratio between H_{2} : O_{2} is always 2:1. 
the ratio between H_{2} : H_{2}0 is always 2:2,
or 1:1. 
the ratio between O_{2} : H_{2}0 is always 1:2. 
This is known as stoichiometric ratio. In unit 7, we learned
that moles and grams can be interconverted by a compound's molar mass (section
7.5.1). By knowing the mole ratio of the substances in the reaction, it enables
us to calculate the mass of the reactants consumed and the mass of products formed.
We must master calculations involving
the different types of stoichiometric problems that arise from a balanced chemical
equation. They are:
 Molemole Calculations (section 8.5.1)
 Massmass Calculations (section 8.5.2)
 Massmole Calculations (section 8.5.3)
 Excess reactant and limiting reagent problems (section
8.5.4)
When you are going through these sections, you need to go slowly and thoroughly to understand the logic in each step. Refer
to your textbook for more examples of each type of stoichiometric problems.
Section
10.1
Information
Obtained from a Balanced Chemical Equation ..p249
