http://www.ck12.org Chapter 12. Stoichiometry
FIGURE 12.2
This image depicts how moles, mass, and
mole ratios are related for a given chemi-
cal equation.0 .500 mol AgCl×(1 mol AgCl1 mol NaCl)(^58 1 mol NaCl.44 g NaCl) = 29 .2 g NaCl
Mass Reactants↔Moles Reactants↔Moles Products↔Mass Products
In the chemistry lab, we frequently need to calculate the relationship between two reactants or products in a chemical
reaction. For example, we may know the mass of one reactant and want to know how much of a given product will
be generated if the reactant is fully consumed. We may also wish to know how much of a second reactant is required
to fully react with the first reactant. These types of questions can be answered by using molar masses and mole ratios
as conversion factors. We will illustrate this process with an example.
Example 12.4
How many grams of lead(II) chloride would be produced if 1.67 g of lead(II) nitrate is allowed to react completely
in the presence of a sodium chloride solution? How many grams of sodium chloride would be consumed in the
process?
Pb(NO 3 ) 2 (aq)+ 2 NaCl(aq)→PbCl 2 (s)+ 2 NaNO 3 (aq)Answer:
First, we need to relate grams of lead(II) chloride to grams of lead(II) nitrate. We can set up the following expression,
using the molar masses of each component and their mole ratio, obtained from the balanced equation:
1.67 Pb(NO 3 ) 2 ×( 331 1 mol Pb.2 g Pb((NONO^3 ) 3 )^22 )(1 mol Pb1 mol PbCl(NO^23 ) 2 )(^278 1 mol PbCl.11 g PbCl 22 )=1.40 g PbCl 2
Therefore, 1.40 g of lead(II) chloride would be produced if 1.67 g of lead(II) nitrate is fully consumed. The amount
of NaCl that would be used in this process can be calculated as follows:
1.67 Pb(NO 3 ) 2 ×( 331 1 mol Pb.2 g Pb(NO(NO^33 )^2 ) 2 )(1 mol Pb2 mol NaCl(NO 3 ) 2 )(^58 1 mol NaCl.44 g NaCl) = 0.589 g NaCl
In order to fully consume 1.67 g of lead(II) nitrate, we would need at least 0.589 g of NaCl.
Lesson Summary
- Using molar masses and mole ratios, we can find the relationships between the masses of various reaction
components for a given reaction.