1) The document discusses quantitative chemistry and limiting/excess reactants. It provides examples of calculations to determine the limiting reactant and amount of product formed based on the limiting reactant. 2) It also discusses percent yield calculations. For the example reaction of gold (III) chloride and lead, the theoretical yield of gold is calculated to be 9.09g based on the given amount of gold (III) chloride. 3) The actual yield obtained in the example is 1.05g gold. The percentage yield is calculated to be 11.6%, expressing the proportion of theoretical yield that was actually obtained.

1. IB Chemistry Power Points
Topic 1
Quantitative Chemistry
www.pedagogics.ca
Lecture 3
Limiting Reactant
Percent Yield

2. Consider the following reaction
2 H2 + O2  2 H2O

3. Reactants are combined in
perfect proportions
3 molecules
6 molecules 6 molecules

4. In reality this never happens
3 molecules
6 molecules 6 molecules

5. Consider
3 molecules
4 molecules 4 molecules +
leftover oxygen

6. Consider
EXCESS
LIMITING REACTANT Amount of
REACTANT PRODUCT is
determined
by limiting
reactant

7. Consider
2 molecules
6 molecules 4 molecules +
leftover
hydrogen

8. Consider
LIMITING
EXCESS REACTANT Amount of
REACTANT PRODUCT is
determined
by limiting
reactant

9. Given 24 grams of O2 and 5.0 grams of H2
determine the mass of H2O produced.
2 H2 + O2  2 H2O
the mass of H2O produced will be determined by the
limiting reactant - do TWO calculations

10. calculation for 24 grams of O2
24 g O2 2 H2O 18.0 g mol-1
= 27 g of H2O
32.0 g mol-1 1 O2

11. calculation for 24 grams of O2
24 g O2 2 H2O 18.0 g mol-1
= 27 g of H2O
32.0 g mol-1 1 O2
calculation for 5.0 grams of H2
5 g H2 2 H2O 18.0 g mol-1
= 45 g of H2O
2.0 g mol-1 2 H2

12. calculation for 24 grams of O2
24 g O2 2 H2O 18.0 g mol-1
= 27 g of H2O
32.0 g mol-1 1 O2
O2 is the LIMITING REACTANT and determines the
amount of product
calculation for 5.0 grams of H2
5 g H2 2 H2O 18.0 g mol-1
= 45 g of H2O
2.0 g mol-1 2 H2
H2 is the EXCESS REACTANT (some would be left
over)

13. How much hydrogen gas would be left over?
To calculate, first determine how much reacts
with all of the oxygen

14. How much hydrogen gas would be left over?
To calculate, first determine how much reacts
with all of the oxygen
given 24 grams of O2
24 g O2 2 H2 2.0 g mol-1
= 3.0 g of H2
32.0 g mol-1 1 O2
3.0 g of H2 reacts so

15. How much hydrogen gas would be left over?
To calculate, first determine how much reacts
with all of the oxygen
given 24 grams of O2
24 g O2 2 H2 2.0 g mol-1
= 3.0 g of H2
32.0 g mol-1 1 O2
3.0 g of H2 reacts so
5.0 g – 3.0 g = 2.0 g of hydrogen remains

16. Percent Yield
Enoch the Red, an alchemist,
wants to try to turn lead into
gold (which you can’t do
chemically). He finds that
mixing lead with an
unidentified compound (gold
III chloride) actually produces
small amounts of gold. The
reaction is as follows:
2 AuCl3 +3 Pb  3 PbCl2 + 2 Au

17. Percent Yield
Enoch reacts 14.0 g of gold III
chloride with excess lead
metal. What would be the
maximum, THEORETICAL yield
of this reaction?
2 AuCl3 +3 Pb  3 PbCl2 + 2 Au

18. Percent Yield
Enoch reacts 14.0 g of gold III
chloride with excess lead
metal. What would be the
maximum, THEORETICAL yield
of this reaction?
2 AuCl3 +3 Pb  3 PbCl2 + 2 Au

19. Percent Yield
Enoch reacts 14.0 g of gold III
chloride with excess lead
metal. What would be the
maximum, THEORETICAL yield
of this reaction?
2 AuCl3 +3 Pb  3 PbCl2 + 2 Au
given 14.0 g of AuCl3
14.0 g AuCl3 2 AuCl3 196.97 g mol-1
= 9.09 g Au
303.5 g mol-1 2 Au

20. Percent Yield
Enoch recovers only 1.05 g of
gold from the reaction. This
could be for many different
reasons
some product was lost in the recovery process
the reaction did not go to completion
the AuCl3 is not pure

21. Percent Yield
the percentage yield expresses
the proportion of the expected
product that was actually
obtained.
actual
% yield= ×100%
theoretical
1.05
% yield= ×100%=11.6%
9.09