2. Grilled Cheese
Sandwich
Bread + Cheese ‘Cheese Melt’
2 B + C B2C
100 bread 30 slices ? sandwiches
What is the limiting factor in our ability
to make the maximum about of grilled
cheese sandwiches containing 2 slices
of bread and 1 slice of cheese?
3. LIMITING REACTANT
IMPORTANCE:
Calculations of limiting reactant bring quantitative
understanding to chemical reactions
These calculations are used in both General and
Organic Chemistry
4. DEFINITIONS
LIMITING REACTANT
Completely consumed in a chemical reaction
Determines the amount of product formed
The reactant that produces the least amount
of product
5. DEFINITIONS
THEORETICAL YIELD
The amount of product that can be made based on
the amount of the limiting reactant
ACTUAL YIELD
The amount of product actually or experimentally
produced
THE PERCENT YIELD
%yield = (actual/theoretical) x 100
6. Limiting Reactants.
An analogous situation occurs with chemical
reactions. Consider the reaction:
2 H2(g) + O2(g) → 2 H2O(l)
2 mol + 1 mol 2 mol
If we have exactly 2 mol of H2 and 1 mol of O2, then
we can make 2 mol of water. But what if we have 4
mol of H2 and 1 mol of O2. Now we can make only 2
mol H2O with 2 mol H2 left over. In this case the O2
is the limiting reagent.
The limiting reagent is the one with nothing left over.
10. Before and After Reaction 2
Before the reaction After the reaction
11. Multiplying an equation through by a
common multiple:
We can multiply all the coefficients in a balanced
equation by any multiple, and it still has the correct
ratios of moles. Thus, if we have:
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
1 mole 2 moles 1 mole + 1 mole
If we have 2 moles of Zn(s), this gives: (x 2)
2 moles 4 moles 2 moles 2 moles
or if we have 0.5 moles Zn(s) we have: (x 0.5)
0.5 moles 1 mole 0.5 moles 0.5 moles
12. METHODS USED TO
DETERMINE THE LIMITING
REACTANT
I. Calculate the moles needed of each reactant and compare with
the moles given
II. Divide the moles of each reactant by its stoichiometric coefficient
and then compare them
III. Calculate the moles of product produced by each reactant and
compare them
13. Example I. Consider the reaction of H2 and N2 to give
NH3, and assume we have 3.0 mol N2 and 6.0 mol
H2.
We have the balanced equation:
N2(g) + 3 H2(g) → 2 NH3(g)
1 mol 3 mol 2 mol
Factor = moles N2 we have
moles N2 in equation
= 3.0 mol N2
1.0 mol N2
= 3.0 (multiply all coefficients in
balanced equation by this factor)
14. Multiply all coefficients by factor (x 3):
N2(g) + 3 H2(g) → 2 NH3(g)
1 mol 3 mol 2 mol
3 mol 3 x 3 = 9 mol 3 x 2 = 6 mol
Try N2 as limiting reagent:
3 mol N2 requires how many moles H2?
= 3 x 3 = 9 mol
We only have 6 mol H2, so H2 is the
limiting reagent.
15. Example II. Divide the moles of each
reactant by its stoichiometric
coefficient
Consider the following reaction:
2 Na3PO4(aq) + 3 Ba(NO3)2(aq) Ba3(PO4)2 + 6
NaNO3
How much Ba3(PO4)2 can be formed if we have in the solutions 3.50 g
sodium phosphate and 6.40 g barium nitrate?
16. Step 1. Convert to moles:
First work out numbers of Moles:
Na3PO4 = 3.50 g x 1 mol = 0.0213 mol
164 g
Ba(NO3)2 = 6.40 g x 1 mol = 0.0245 mol
261 g
18. Example III. Calculate the amount of product produced by each reactant
1N2(g) + 3H2(g) → 2NH3(g)
Given 3.0 mole 6.0 mole
3.0 mol N2 x 2 mol NH3 = 6.0 mol NH3
1 mol N2
6.0 mol H2 x 2 mol NH3 = 4.0 mol NH3 (theoretical yield)
3 mol H2
The reactant that produces the least amount of product is the
L.R.H2
19. Practice Exercise:
Zn metal (2.00 g) plus solution of AgNO3 (2.50 g) reacts according to:
Zn(s) + 2 AgNO3(aq) Zn(NO3)2+ 2 Ag(s)
1 mol 2 mol
Which is the limiting reagent?
How much Zn will be left over?
20. Step 1. Convert to moles:
Zn = 65 g/mol
AgNO3 = 108 + 14 + (3 x 16) = 170 g/mol
Zn = 2.0 g x 1 mol= 0.0308 mol
65 g
AgNO3 = 2.50 g x 1 mol = 0.0147 mol
170 g
21. Step 2. Guess limiting reagent
Zn(s) + 2 AgNO3(aq) Zn(NO3)2+ 2 Ag(s)
1 mol 2 mol
0.0308 0.0147
In this case it seems clear that AgNO3 must be the
limiting reagent, because the equation says we
must have 2 mols of AgNO3 for each mol of Zn(s),
but in fact we have more moles of Zn(s).
22. We can check this by dividing the moles of
each reactant by their coefficients
AgNO3 = 0.0147/2 = 0.00735
Zn = 0.0308/1 = 0.0308
Zn(s) + 2 AgNO3(aq) Zn(NO3)2+ 2 Ag(s)
1 mol 2 mol
0.0308 0.00735
We in fact have 0.0305 mol of Zn, which is
more than the 0.00735 mol of AgNO3, so
AgNO3 is clearly the limiting reactant.
23. How much Zn is left over?
Use the limiting reactant to determine this:
0.0147 mol AgNO3 x 1 mol Zn x 65 g Zn
2 mol AgNO3 1 mol Zn
= 0.47775 g Zn
Subtract this from the amount of Zn available:
2.00 g Zn - 0.4775g Zn = 1.52 g Zn in excess
24. Percent Yield:
Theoretical yields:
The quantity of product that forms if all of the limiting reagent
reacts is called the theoretical yield. Usually, we obtain less than this,
which is known as the actual yield.
Percent yield = actual yield x 100
Theoretical yield
25. Problem:
10.4 g of Ba(OH)2 was reacted with an excess of
Na2SO4 to give a precipitate of BaSO4. If the reaction
actually yielded 11.2 g of BaSO4, what is a) the
theoretical yield of BaSO4 and b) what is the
percentage yield of BaSO4?
The balanced equation for the reaction is:
Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaOH(aq)
26. Step 1. Convert to moles:
Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaOH(aq)
1 mole 1 mole 1 mole 2 moles
Moles Ba(OH)2:
Mol. Mass Ba(OH)2 = 137.3 + 2 x (16.0 + 1.0)
= 171.3 g/mol
Moles = 10.4 g x 1 mol = 0.0607 moles
171.3 g
27. Step 2. Work out how much BaSO4
will be formed:
Ba(OH)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaOH(aq)
1 mole 1 mole 1 mole 2 moles
0.0607 moles 0.0607 moles
When it says that one reagent is in excess, that means we
do not have to worry about that reagent, and the other
one is the limiting reagent, in this case the BaSO4.
We see that 1 mole of Ba(OH)2 will produce 1 mole of
BaSO4. Our factor is thus 0.0607, and we will get 0.0607
moles of BaSO4.
28. Convert actual yield to percentage yield:
Percent yield = actual yield x 100 %
Theoretical yield
= 11.2 g x 100 %
14.29 g
= 78.4 % yield
29. A student placed 5.00 g Na in a flask and added
enough Cl2 for the Na to react completely. The
student collected 10.00 g NaCl. What percent yield
did the student achieve? (Which means how much
product did the student collect as compared to the
theoretical maximum.) So, % yield = amt. isolated x
100 theoretical amt. What amounts? Grams (it is
possible to use grams, but most people get very
confused when they try to use grams)? No! compare
moles. (Actually, the important thing is to always
compare like units. Compare moles to moles, or
grams to grams, but never grams to moles.)
30. To compare moles first write the balanced equation.
2 Na + Cl2 ———> 2 NaCl
How many moles of Na of Na did the student use?
5.00 g x 1 mol Na = 0.2175 mol Na
22.99 g Na
So, how many moles of NaCl should the student be able to make?
0.2175 mol Na x 2 mol NaCl = 0.2175 mol NaCl (theoretical
limit)
2 mol Na
So, we know how much the student could have made, but
how much did the
student make?
10.00 g NaCl x 1 mole NaCl = 0.17110 mol NaCl (isolated)
58.443 g NaCl
So, 0.17110 mol NaCl (collected) x 100 = 78.7 % yield
0.2175 mol NaCl (theoretical
31. 12.5 g of sodium sulfate and 35.0 g of barium nitrate react to
form
sodium nitrate and barium sulfate. 9.5 g of barium sulfate
were
collected. Determine the percent yield of barium sulfate.
Na2SO4+ Ba(NO3)2 ----------2 NaNO3 + BaSO4
To determine % yield the actual yield and the theoretical
maximum yield must be known.
Actual yield = 9.5 g barium sulfate
32. The reagent that limits the reaction has to be determined in order
to determine the theoretical
yield.
Sometimes it is just easier to determine the theoretical yield
possible from each reagent, and the lower yield is the theoretical
yield.
35.0 g Ba(NO3)2 1 mol Ba(NO3)2 x 1 mol BaSO4
261.34 g Ba(NO3)2 1mol Ba(NO3)2 = 0.134
mol BaSO4
There is enough Ba(NO3)2 to make 0.134 mol BaSO4.
12.5 g Na2SO4 x 1 mol Na2SO4 x 1 mol BaSO4
142.04g Na2SO4 1mol Na2SO4 = 0.0880 mol
BaSO4
However, there is only enough Na2SO4 to make 0.088 mol
BaSO4
33. Once the supply of Na2SO4 is exhausted the reaction
will stop.
Theoretical yield is 0.0880 mol BaSO4
To determine % yield you must compare the same
units; i.e., compare moles to moles, or grams to
grams, but never grams to moles.
% yield = actual yield x 100
theoretical yield
Since the actual yield is reported in grams that
number must be converted to moles.
35. QUIZ # 1
1. Consider the following reaction:
NH4NO3 + Na3PO4----->( NH4)3PO4 + NaNO3
Which reactant is limiting, assuming we started
with 30.0 grams of ammonium nitrate and 50.0
grams of sodium phosphate. a. Balance the
reaction b. What is the mass of each product
that can be formed? c. What mass of the
excess reactant(s) is left over?
Ans.18.6 grams of ammonium phosphate, 31.9
grams of sodium nitrate 29.5 grams of sodium
phosphate
36. The equation for the reaction of iron (III) phosphate with
sodium sulfate to make iron (III) sulfate and sodium
phosphate.
2 FePO4 + 3Na2SO4 ---> Fe2(SO4)3 + 2 Na3PO4
a. a) If I perform this reaction with 25 grams of iron (III)
phosphate and an excess of sodium sulfate, how many
grams of iron (III) sulfate can I make?17.2g
b. If 18.8 grams of iron (III) sulfate are actually made when
I do this reaction, what is my percent yield? 109.3
c. Is the answer from problem b) reasonable? Explain.
d. If I do this reaction with 15 grams of sodium sulfate
and get a 65.0% yield, how many grams of sodium
phosphate will I make? According to the stoichiometry,
the theoretical yield is 11.5 grams. Multiplying this by
0.650, you get 7.48 grams.
37. ASSIGNMENT 1:
Consider the following balanced reaction:
3CaCO3+2FePO4 -->Ca3(PO4)2+Fe2(CO3)3
Which reactant is limiting, assuming we start
with 100. grams of calcium carbonate and 45.0
grams of iron (III) phosphate. a. What is the
mass of each product that can be formed?
b. What mass of the excess reactant(s) is left
over?
