Page 155 - Theory and Problems of BEGINNING CHEMISTRY

144 STOICHIOMETRY [CHAP. 10

Since the mole ratio is given by the equation, we must convert 124 g of NaOH to moles:

1 mol NaOH
124 g NaOH = 3.10 mol NaOH
40.0 g NaOH
Now we can solve the problem just as we did those above:

1 mol Na 2 SO 4
3.10 mol NaOH = 1.55 mol Na 2 SO 4
2 mol NaOH
The left two boxes of Fig. 10-2 illustrate the additional step required for this calculation.
Molar Balanced chemical Molar
Mass A Moles A Moles B Mass B
mass equation mass
Fig. 10-2. Conversion of mass of a reactant to mass of a product
It is also possible to calculate the mass of product from the number of moles of product.

EXAMPLE 10.6. How many grams of Na 2 SO 4 can be produced from 124 g NaOH with sufficient H 2 SO 4 ?
Ans. The first steps were presented in Example 10.5. It only remains to convert 1.55 mol Na 2 SO 4 to grams (Fig. 10-2):

142gNa SO 4
2
1.55 mol Na 2 SO 4 = 220gNa SO 4
2
1 mol Na 2 SO 4
Not only mass, but any measurable quantity that can be converted to moles may be treated in this manner
to determine the quantity of product or reactant involved in a reaction from the quantity of any other reactant or
product. (In later chapters, the volumes of gases and the volumes of solutions of known concentrations will be
used to determine the numbers of moles of a reactant or product.) We can illustrate the process with the following
problem.

EXAMPLE 10.7. How many grams of Na 3 PO 4 can be produced by the reaction of 1.11 × 10 23 formula units of NaOH
with sufficient H 3 PO 4 ?
Ans. 3 NaOH + H 3 PO 4 −→ Na 3 PO 4 + 3H 2 O
Since the equation states the mole ratio, first we convert the number of formula units of NaOH to moles:
1 mol NaOH
23
1.11 × 10 formula units NaOH = 0.184 mol NaOH
6.02 × 10 formula units NaOH
23
Next we convert that number of moles of NaOH to moles and then grams of Na 3 PO 4 :

1 mol Na 3 PO 4
0.184 mol NaOH = 0.0613 mol Na 3 PO 4
3 mol NaOH

164gNa PO 4
3
0.0613 mol Na 3 PO 4 = 10.1gNa PO 4
3
1 mol Na 3 PO 4
Figure 10-3 illustrates the process.
Avogadro’s Balanced Molar
Formula Moles A Moles B Mass B
units A number chemical mass
equation
Fig. 10-3. Conversion of number of formula units of a reactant to mass of a product

10.3. LIMITING QUANTITIES
In Secs. 10.1 and 10.2, there was always sufficient (or excess) of all reactants except the one whose quantity
was given. The quantity of only one reactant or product was stated in the problem. In this section, the quantities
of more than one reactant will be stated. This type of problem is called a limiting-quantities problem.