Page 156 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 10] STOICHIOMETRY 145

How much Al 2 O 3 can be prepared from 2.0 mol of O 2 and 0.0 mol of Al? The first step, as usual, is to write
the balanced chemical equation:

4Al + 3O 2 −→ 2Al 2 O 3

It should be obvious that with no Al, there can be no Al 2 O 3 produced by this reaction. (This problem is not one
which is likely to appear on examinations.)
How much sulfur dioxide is produced by the reaction of 1.00 g S and all the oxygen in the atmosphere of
the earth? (If you strike a match outside, do you really have to worry about not having enough oxygen to burn
all the sulfur in the match head?) This problem has the quantity of each of two reactants stated, but it is obvious
that the sulfur will be used up before the oxygen. It is also obvious that not all the oxygen will react! (Otherwise,
we are all in trouble.) The problem is solved just as the problems in Sec. 10.2.
To solve a limiting-quantities problem in which the reactant in excess is not obvious, do as follows: If the
reagents appear in a 1 mol: 1 mol ratio in the balanced chemical equation, you can tell immediately that the
reagent present in lower number of moles is the one in limiting quantity. If they are not in a 1 mol: 1 mol ratio, an
easy way to determine which is in limiting quantity is to divide the number of moles of each by the corresponding
coefficient in the balanced chemical equation. The reagent with the lower quotient is in limiting quantity. Do not
use these quotients for any further calculations. (In fact, it is useful to draw a line through them as soon as you
see which reactant is in limiting quantity.) We can use the reactant in limiting quantity to determine the number
of moles of each product that will be produced and the number of moles of the other reactant(s) that will be used
up in the reaction. These steps are illustrated in Example 10.8 below.
In a limiting-quantities problem, you might be asked the number of moles of every substance remaining
after the reaction. A useful way to calculate all the quantities is to use a table to do the calculations.

Step 1: Use the balanced equation as the heads of the quantities in the table, and label the rows “Present initially,”
“Change due to reaction,” and “Present finally.” The initial quantities of reactants are entered in the first row.
Zero is entered under the heading for any substances not present.
Step 2: Start the second row by entering the moles of limiting quantity—the same as the number of moles of
that substance in the first row, since all of the limiting quantity is used up. The other entries in the second row
are calculated using the technique of Sec. 10.1. The entries in row 2 will always be in the same mole ratio as the
coefficients in the balanced chemical equation.
Step 3: Calculate the third row entries by adding the numbers of moles of products in the first two rows and
subtracting the numbers of moles of each reactant in the second row from the corresponding number in the first
row. (If you ever get a negative number of moles in row 3, you have made a mistake somewhere.)

EXAMPLE 10.8. Calculate the number of moles of each substance present after 4.60 mol of aluminum is treated with
4.20 mol of oxygen gas and allowed to react.
Ans. The balanced equation and the left table entries are written first, along with the initial quantities from the statement
of the problem:
4Al + 3O 2 −→ 2Al 2 O 3
Present initially: 4.60 mol 4.20 mol 0.00 mol
Change due to reaction:
Present finally:
Dividing 4.60 mol of Al by its coefficient 4 yields 1.15 mol Al; dividing 4.20 mol O 2 by 3 yields 1.40 mol
O 2 . Since 1.15 is lower than 1.40, the Al is in limiting quantity. Therefore 4.60 mol is entered in the second
row under Al. Note that the number of moles of Al present is greater than the number of moles of O 2 ,but
the Al is still in limiting quantity. The quantity of O 2 that reacts and of Al 2 O 3 that is formed is calculated as
in Sec. 10.1:

3 mol O 2 2 mol Al 2 O 3
4.60 mol Al = 3.45 mol O 2 4.60 mol Al = 2.30 mol Al 2 O 3
4 mol Al 4 mol Al