KEY Unit 6 - Reaction Stoichiometry and Three Types of
Reaction in Aqueous Solution
Limiting Reactant and Theoretical Yield
Limiting Reagent (or reactant) - the reactant that is completely consumed during a
chemical reaction and limits the amount of product which is formed.
Theoretical Yield - the maximum amount of product which can be formed if all of the
limiting reagent is consumed and the reaction proceeds to completion.
For example:
Based on the following reaction, how many grams of the product (POCl3 (s)) can be
formed starting with 10.0 grams of each of the reactants?
6 PCl3 (l) + 6 Cl2 (g) + P4O10 (s) ===> 10 POCl3 (s)
10.0 g
10.0 g
10.0 g
??? g
Convert all initial quantities into mole units using the correct conversion factors (in
this case molar masses).
10.0 g
= 0.0728 2 mole
g
137.33
mole
10.0 g
moleCl2 =
= 0.1410 mole
g
70.906
mole
10.0 g
moleP4O10 =
= 0.0352 3 mole
283.89 g
mole
molePCl3 =
Determine the limiting reactant using sound chemical reasoning and calculations. For
example, in this case we might state:
“The chlorine gas cannot be limiting because it is consumed in the same molar
quantity as is the phosphorus trichloride but is present in about double the quantity.
The choice then is between the phosphorus trichloride and the tetraphosphorus
decoxide. Since 0.0728/6 is less than 0.0352 it is clear that we will run out of the PCl3
long before the P4O10 is entirely consumed. ” . . . So PCl3 is the limiting reactant (LR).
Find the theoretical yield of product formed by the LR.
gPOCl3 =
153.33 gPOCl3 10molePOCl3
x
x0.07282molePCl3 = 18.61 gPOCl3
1molePOCl3
6molePCl3
Ans.: 18.6 g of POCl3 (s)
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Experimental Yield: Percent Yield
Even if all reactants in a reaction are at stoichiometry, that reaction often does not go to
completion. This can result from:
•
•
•
•
unfavorable conditions
reaction being stopped before it was completed
side reactions
back reaction leading to equilibrium
Because of the above considerations, the concept of reaction yield is quite important.
Actual Yield - the experimentally measured yield of product of a reaction.
Percent Yield - the ratio of actual yield to theoretical yield expressed as a percentage.
%Yield =
actual yield
x100
theoretical yield
Ex:
(a) What if the reaction actually yields 18.5 g POCl3 (s)?
(b) What if the reaction yields only 13.4 g POCl3 (s)?
Ans.:
(a) 99.5 % [see equation above in plug in numbers!!!!]
(b) 72.0 %
YOU CAN ALSO TRY THESE:
(1) 10.0 g of aluminum (Al) reacts with 50.0 g of sulfur (S) to produce aluminum sulfide
(I'll let you determine the formula of this compound and write the balanced chemical
equation).
The reaction may be written in two forms (both will yield the same answers – provided
the proper conversion factors are used:
2 Al (s) + 3 S (s) Æ Al2S3 (s) or
10.0 g
50.0 g
16 Al (s) + 3 S8 (s) Æ 8 Al2S3 (s)
(a) What is the limiting reactant?
for Al we find 10.0 g = 0.3706 mole Al
for S we find 50.0 g = ether 1.559 mole S
or 0.1949 mole S8
therefore the limiting reactant is Al (s) b/c 1.5*(0.3706)<1.559
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(b) What is the theoretical yield of aluminum sulfide?
We get one mole of aluminum sulfide for every two moles of aluminum metal
So we find that we can produce a maximum of 0.1853 mole Al2S3 (s) . . .
and this corresponds to 27.82 g Al2S3 (s)
(c) How many grams of the nonlimiting reactant would react and how much
would remain unreacted?
If 27.82 g of Al2S3 (s) contains 10.0 g of Al, then it must contain 17.82 g S. Which
means that (50.0 – 17.82)g = 32.18 g of S must remain. This calculation relies on
mass conservation. We could also find the number of moles of S consumed (1.5
times the number of moles of Al consumed) and then converting that to mass
units of sulfur.
(
3mole S
/2 mole Al)*(0.3706 mole Al) = 0.5559 mole S = 17.82 g S
(d) Verify mass conservation.
see (c) above and find that:
50.0 g S + 10.0 g Al yields 27.82 g Al2S3 + 32.18 g S (or S8) “left-overs”
(e) What is the % yield if the actual yield is 25.0 g of aluminum sulfide?
25.0 g
% yield = 100*(
/27.82 g) = 89.86 %
Ans.:
(a) Al
(b) 27.8 g
(c) 17.8 g S react and 32.2 g does not react
(d) 10.0 g + 50.0 g = 27.8 g + 32.2 g
(e) 89.9 %
(2) Given the following reaction, find the theoretical yield of both products. Also,
calculate the % yield if 5.92 g of Pt(NH3)2Cl2 is actually formed. Based on this % yield,
how much KCl would you expect was actually formed?
K2PtCl4 (s) + 2 NH3 (g) ===> 2 KCl (s) + Pt(NH3)2Cl2
8.50 g
1.00 g
??? g
??? g
Ans.: follow the same type of logic employed above to find that
3.06 g of KCl and 6.15 g of Pt(NH3)2Cl2
96.3 % yield
2.95 g KCl
Quantitative Solution Concentrations
You can try these:
1. (a) How many grams of NaCl are needed to prepare 5.00 x 102 mL of 0.250 M NaCl?
(b) What volume of this solution will contain 0.050 moles of NaCl?
(c) What volume of 0.250 M NaCl is needed to prepare 75.0 mL of 0.150 M NaCl?
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Ans.:
58.44 g NaCl
/ 1 mole NaCl)(0.250 mole/Lsol’n)(0.500L)
(a) 7.31 g of NaCl = (
(b) 2.00 x 102 mL of 0.250 M NaCl = (
(c) 45.0 mL of 0.250 M NaCl =
1000 mL sol’n
/ 0.250 mole NaCl)*(0.050 mole)
(75.0 mL*0.150M)
/(0.250 M)
Solution Stoichiometry
We can now determine the number of moles of a substance from information about its
concentration in a solution and the volume of the solution. This gives us a new way to get
information about limiting reactants in reaction stoichiometry problems.
For example, determine the maximum mass of iron (III) carbonate precipitate that can be
produced from the quantities of reactants shown below.
3 Na2CO3 (aq) + 2 Fe(NO3)3 (aq) ===> Fe2(CO3)3 (s) + 6 NaNO3 (aq)
200.0 mL
200.0 mL
??? g
0.500 M
0.500 M
Helpful info.: The molar mass of Fe2(CO3)3 (s) = 291.7 g/mole
Ans.: 9.72 g Fe2(CO3)3 (s)
(200.0 mL)(0.500 M) = 100 mmoles sodium carbonate = 0.100 mole
(200.0 mL)(0.500 M) = 100 mmoles iron (III) nitrate = 0.100 mole
since we require only 2 moles the iron reagent for every 3 moles of the carbonate source,
the carbonate will run out before all of the iron is precipitated
291.7 g iron (III) carbonate
mass ppt = (
mass ppt = 9.72 g
/ 1 mole) (1mole ppt/ 3 mole carbonate)*0.100 mole
As a follow-up question we could very well ask . . . How many mL of 0.500 M Fe(NO3)3
are required to completely react with 200.0 mL of 0.500 M Na2CO3 (aq)?
(aq)
Ans.: 133 mL Fe(NO3)3 (aq) = . . .
(1000 mL iron sol’n/0.500 mole Fe3+)*(2 mole Fe/3 mole carbonate)* 0.100 mole carbonate
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Ex.: How many mL of 0.200 M NaOH(aq) is required to completely neutralize:
(a) 40.0 mL of 0.0500 M H2SO4 (aq)
(b) 40.0 mL of 0.0500 M H3PO4 (aq)
Ans.:
(a) 20.0 mL of 0.200 M NaOH(aq)
write balanced chemical equation showing that you need 2 NaOH for each H2SO4
2 NaOH (aq) + H2SO4
(aq)
Æ 2 H2O (l) + Na2SO4
(aq)
then do the math to find the mmoles of acid to be neutralized
(0.0500 mmole H2SO4/1mL)*40.0mL = 2.00 mmole of acid
and the moles and mL of base needed to get the job done
mmole of base =
mL of base = (
(2 mmole NaOH/1 mmole H2SO4)*2.00 mmole acid = 4.00 mmole
1mL NaOH
/0.200 mole)*4.00 mmole NaOH = 20.0 mL of the base
(b) 30.0 mL of 0.200 M NaOH(aq)
write balanced chemical equation showing that you need 3 NaOH for each H3PO4
3 NaOH (aq) + H3PO4
(aq)
Æ 3 H2O (l) + Na3PO4
(aq)
then do the math to find the mmoles of acid to be neutralized
(0.0500 mmole H3PO4/1mL)*40.0mL = 2.00 mmole of acid
and the moles and mL of base needed to get the job done
mmole of base =
mL of base = (
(3 mmole NaOH/1 mmole H3PO4)*2.00 mmole acid = 6.00 mmole
1mL NaOH
/0.200 mole)*6.00 mmole NaOH = 30.0 mL of the base
Ex.: 35.67 mL of 0.1000 M HCl(aq) is required to neutralize 25.00 mL of a NaOH(aq)
solution of unknown concentration. What is [NaOH]? Ans.:
Ans: [NaOH] = 0.1427 M
I’ll leave this one to the students . . . after all, you’ve done this in the lab this semester.
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