Note that we proceeded in five steps:
1.Write
an unbalanced expression with the correct reactants and products.
2.Balance
the equation properly, and obtain the ratio of number of moles of
the reactants and products of interest.
3.Calculate
molecular weights of the reactants and products of interest.
4.Calculate
the number of moles of reactant used, and use the mole ratio from
step 2 to find the number of moles of product.
5.Use the
molecular weight of the product to obtain the weight in grams.
Example.
Glucose is a sugar with the chemical formula ,
and is a common energy source for living organisms. How many moles
of oxygen are required to burn a mole of glucose, and how many grams
of
are needed for a kilogram of glucose?
In passing, observe that burning a kilogram of glucose requires
only 1066 g of oxygen, whereas burning the same weight of propane
requires 3627 g of oxygen. This is because glucose already is partially
oxidized. You should not be surprised later when we calculate that
the combustion of glucose produces only half as much heat per gram
as combustion of propane. Glucose is a poorer fuel than propane
on a weight basis. In Chapter 21 we will return to the question
as to why plants selected glucose for energy storage (in plant starch),
whereas animals developed fats as a kind of "solid propane."