Chemistry Reference
In-Depth Information
Practice Problem 10.27
Calculate the percent yield if 14.1 g of
is obtained from treatment of 10.0 g of
PCl
5
PCl
3
with 6.00 g of gaseous Cl
2
.
EXAMPLE 10.28
Ozone, is produced when molecules are subjected to electrical discharge
or the action of cosmic rays in the upper atmosphere.
(a)
O
3
,
O
2
Calculate the mass of ozone that could theoretically be produced by
conversion of 1.47 g of
O
2
.
(b)
If 0.111 g of
O
3
is actually produced, what is the percent yield?
Solution
(a)
3 O
2
(g) £ 2 O
3
(g)
a
1 mol O
2
32.0 g O
2
a
2 mol O
3
3 mol O
2
a
48.0 g O
3
1 mol O
3
1.47 g O
2
b
b
b
1.47 g O
3
This part of the problem could have been solved by simply applying the
law of conservation of mass.
(b)
The percent yield is
a
0.111 g
1.47 g
b
100%
7.55%
yield
Snapshot Review
❒
The calculated quantity of product (Sections 10.1-10.4) is the largest
possible quantity that can ever be expected, but many factors may
reduce the actual yield of product. Competing reactions, solute
remaining in solution, and reactions not going to completion
(Chapters 18 and 19) all lead to less product than calculated. The
percent yield is 100% times the ratio of the actual yield divided by
the theoretical yield.
ChemSkill Builder 4.4
A. If 12.1 g of a certain reactant is calculated to yield 15.5 g of a product,
but only 10.9 g of product is obtained, what is (a) the theoretical yield?
(b) the actual yield? (c) the percent yield?
10.6
Calculations with Net Ionic Equations
Net ionic equations (Chapter 9), like all other balanced chemical equations, give
the mole ratios of reactants and products. Therefore, any calculations that require
mole ratios may be done with net ionic equations as well as with total equations.
However, a net ionic equation does not yield mass data directly because part of
each soluble ionic compound is not given. For example, we can tell how many
moles of silver ion are required to produce a certain number of moles of a product,
