NAME ___________________________ NOTES: UNIT 4 (5): LIMITING REAGENT & PER CENT YIELD
(This is Purely Honors Chemistry...See p.368 – 377 of the text)
First, a little vocabulary. You might have noticed that in class I have used the terms: reactant & reagent
interchangeably. The terms reactant and reagent essentially are synonyms of each other.
Secondly, have you noticed that the stoichiometry practice problems often contain the phrase “in an excess of...”?
For instance:
Given: 2 Ag2S(s) + 3 O2(g)  2 SO2(g) + 2 Ag2O(s)
Calculate the mass, in grams of Ag2O(s) produced when 7.85 grams of Ag2S(s) are consumed
completely, in an excess of dioxygen.
This phrase tells the reader that there is so much oxygen available that the silver sulfide will be consumed completely
and the reaction ends with left-over dioxygen in the vessel.
Translation: The mass of product is limited by the available amount of reactant(s), as linked to the mole ratios
balanced equation. Since compounds are formed in definite proportions of their elements, you
may “run out of one reactant” and have an excess of others, as the chemical reaction ends. In this
example, the masses of SO2(g) & Ag2O(s) that can be produced are limited by the limited mass of
Ag2S(s) …not the dioxygen, since the dioxygen is in excess
Think about it this way: You’re making peanut butter and jelly sandwiches (mmmmm!) Assume the term
“sandwich” means 2 pieces of bread, two dips of the knife into the peanut butter and 1 dip into the jelly.
You have a 1 lb jar of PB, a 1 lb jar of “J” with four pieces of bread. What is the limiting reagent?
Well, the reactant (reagent) of a chemical reaction which gets used up first is called the limiting reagent. This has big
implications, for it is the concentration of the limiting reagent in relation to the mole ratios of the balanced equation which
determines the amount of product(s) produced.
METAPHOR!! A chain is only as strong as its weakest link. … You can only produce as much
product as you have reactants in the stoichiometric amounts. When one or more of the
reactant concentrations is/are limited, either due to mass or large requirement (as given
by the molar ratio of the coefficients of the balanced equation), then a reaction may not
“go to completion”. The amount of product may be limited by the reactant.
Given:
2 Na(s) + Cl2(g)  2 NaCl(s)
How many moles of of sodium chloride that can be prepared by the reaction of:
i) 0 mol of Na and 1 mol of Cl2 * no NaCl can be made
Thought Experiment
ii) 2 mol of Na and 1 mol of Cl2 * 2 mol NaCl per bal. equation
iii) 2 mol Na and 2 mol of Cl2
* 2 mol of NaCl ... all the Na would be consumed
and 1 mol of Cl2 is leftover bringing
us back to the situation in i) ... thus
Na is the limiting reagent.
293
VI) Limiting Reagent Problems:
A) Recognizing limiting reagent problems: The problem will have quantities for 2 or more different
reactants ... The reactants will not "completely react".

Recognition: Given: Al4C3(s) + 12 H2O(l)  4 Al(OH)3(s) + 3 CH4(g)
Determine the limiting reagent, when 156 grams of Al4C3(s) are allowed to react with
192 grams of water.
(note, you may solve for the mass of either product)

Recognition:
Given: Al4C3(s) + 12 H2O(l)  4 Al(OH)3(s) + 3 CH4(g)
Calculate the maximum mass, in grams, of methane, that can be produced, when
156 grams of Al4C3(s) are allowed to react with 192 grams of water. (Note, in this case, you
must solve for the mass of CH4)

Recognition:
Given: Al4C3(s) + 12 H2O(l)  4 Al(OH)3(s) + 3 CH4(g)
Calculate the mass of each reactant left in the reaction vessel, after 156 grams of Al4C3(s)
have been allowed to react with 192 grams of water. (It is assumed that the limiting reagent is at
a mass of 0 grams … thus this question is seeking the mass of the remaining (un-reacted) reagent, which is
in excess! This means you must determine the limiting reagent, and then determine how much of the other
reactant is used in the consumption of the limiting reagent or production of product, then find unreacted
mass by subtraction from the given.)
B) Strategy: Ultimately, you will do 2 double mole map stoichiometry problems. Use the quantity of
one reactant as the given, and solving for an amount of only 1 of the products (or the
identified product). Then repeat the process, using the quantity of the other reactant, and
solving for the quantity of the same product.
Whichever calculation results in the smallest amount of product, identifies the limiting
reagent.
To determine how much of a reactant is left over, you could determine how much of it
was used to make the maximum amount of the product, and subtract the mass of the
reactant consumed, from the "given" quantity.
Example: Given: 4 FeS2 + 11 O2  2 Fe2O3 + 8 SO2
When given 951.2 grams of FeS2 and 704.0 grams of O2, determine which reactant
is the limiting reagent.
Run the stoichiometries. Use a reactant value as the "given". You can use either
product as the "desired" ...but use it for both stoichiometric calculations.
calculation 1: g SO2 = 951.2 g FeS2 |1 mol FeS2 | 8 mol SO2 | 64 g SO2 | = 1,015 g SO2
120 g FeS2 4 mol FeS2 1 mol SO2
calculation 2: g SO2 = 704.0 g O2 |1 mol O2 | 8 mol SO2 | 64 g SO2 | = 1,024 g SO2
32 g O2 11 mol O2 1 mol SO2
Since the calculation, using the mass of FeS2 results in the smaller mass of sulfur dioxide, then
FeS2 is the limiting reagent. This would hold true, even if you were to run the stoichiometric calculations
solving for a mass of produced Fe2O3.
294
C) Practice (Basic)
1) Given: 2 Ag2S(s) + 3 O2(g)  2 SO2(g) + 2 Ag2O(s)
Determine the limiting reagent when 532.0 grams of silver sulfide and 201.6 Liters of
dioxygen gas are allowed to react, at STP (Notice, the phrase "reacted completely" is missing and you already
*gramsSO2 = 532.0gAg2S|1 mol Ag2S| 2 mol SO2 |64 g SO2| =137.3 gSO2
248 g Ag2S 2mol Ag2S 1 mol SO2
*grams SO2 = 201.6 L| 1 mol O2| 2 mol SO2 | 64 g SO2| = 384.0 grams SO2
22.4 L O2 3 mol O2 1 mol SO2
You should note that the limiting reagent is not necessarily linked to, the numerical value of the lesser “given” value
The limiting reagent is determined by the given quantity AND the mole ratio between the reactants.
2) Given:
C3H8(g) + 5 O2(g) 
3 CO2(g) + 4 H2O(l)
Determine the limiting reagent when 16.8 L of propane and 16.8 L of dioxygen gas are
mixed in a reaction vessel and allowed to react, according to the ratios of the above balanced
reaction equation, at STP.
* g CO2 = 16.8 L propane | 1 mol propane | 3 mol CO2 | 44 grams CO2|
22.4 L propane 1 mol prop 1 mol CO2
= 99.0 grams CO2
* g CO2 = 16.8 L dioxygen | 1 mol O2 | 3 mol CO2 | 44 grams CO2| = 19.8 grams CO2
22.4 L O2 5 mol O2
1 mol CO2
*dioxygen gas is the limiting reagent and thus gives some deeper understanding as to why
limiting the amount of dioxygen in a fire, may help extinguish the fire … The molar
requirement for dioxygen in a combustion is often much greater than the fuel’s
3) Given:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Calculate the maximum mass, in grams, of CO2 produced given the above reaction when
19.6 g of methane 85.0 g of dioxygen react freely with each other.
* g CO2 = 19.6 g CH4 | 1 mol methane | 1 mol CO2 | 44 grams CO2|
16 g methane 1 mol methane 1 mol CO2
= 53.9 grams CO2
* g CO2 = 85.0 g dioxygen | 1 mol O2 | 1 mol CO2 | 44 grams CO2| = 58.4 grams CO2
32 g O2
2 mol O2
1 mol CO2
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4) Given: Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
Calculate the maximum mass of silver metal produced, when 58.3 grams of copper and
85.0 g of silver nitrate are allowed to react according to the above balanced reaction
* g Ag = 58.3 g Cu | 1 mol Cu | 2 mol Ag | 108 g Ag | = 197 g Ag
64 g
1 mol Cu
1 mol
* g Ag = 85.0 gAgNO3 | 1 mol AgNO3 | 2 mol Ag
| 108 g Ag | = 54.0 g Ag
170 g
2 molAgNO3
1 mol
5) Given: 4 BCl3 + P4 + 6 H2  4 BP + 12 HCl
232 grams of BCl3 were reacted with 155 grams of phosphorus and 10.0 grams of dihydrogen
gas. What’s the maximum mass of BP produced?
* gBP = 232 gBCl3 | 1 mole BCl3 | 4 mol BP | 42 grams BP | = 84.0 grams
116 g BCl3
4 mol BCl
1 mol BP
* g BP = 155g P4 | 1 mole P4 | 4 mol BP
124 g
1 mol P4
| 42 grams BP | = 210. grams
1 mol BP
* gBP = 10.0 g H2| 1 mole H2 | 4 mol BP
2g
6 mol H2
| 42 grams BP | = 140. grams
1 mol BP
*Because it produces the fewest moles of BP…. BCl3 is the limiting reagent, thus, only 84.0 grams of
BP may be produced …regardless of the fact that masses of H2 and P4 remain unreacted
6) Calculate the maximum number of grams of aluminum iodide produced when120.5 grams
of aluminum metal are reacted with 543.1 grams of iodine according to the balanced equation:
NB: The problem does not say “are reacted completely”. This is one clue telling you to approach the problem as a limiting reagent problem
2 Al(s) + 3 I2(s)

2 AlI3(s)
* grams of AlI3 = 120.5 g Al | 1 mol Al | 2 mol AlI3| 408mol AlI3| =
27 grams
2 mol Al
1 mol AlI3
*
1,821 g AlI3
grams of AlI3 = 543.1 g I2 | 1 mol I2 | 2 mol AlI3| 408mol AlI3| = 581.6 g AlI3
254 g I2 3 mol I2
1 mol AlI3
296
7) The complete combustion of ethanol (C2H5OH) is represented by the following reaction.
C2H5OH(l) + 3 O2(l)  2 CO2(g) + 3 H2O(g)+ 1,236 kJ
a) What is the limiting reagent in the reaction when 77.0 grams of ethanol and
19.5 grams of O2 are allowed to react. NB: The problem does not say “are reacted completely”.
* grams CO2 = 19.5 gO2 | 1mole O2| 2 mol CO2 | 44 grams CO2| = 17.9 grams
32 g
3 mol O2
1 mole CO2
* grams CO2 = 77.0 g eth | 1mole eth | 2 mol CO2 | 44 grams CO2| = 147 grams
46 g eth
1 mol eth
1 mole CO2
b) What reagent is in excess? * ethanol or C2H5OH
8) Given: 2 N2(g) + 3 O2(g)  2 N2O3(g)
a) Identify the limiting reagent, given 14.0 grams of dinitrogen gas and 96.0 grams of
dioxgyen gas.
* moles N2O3(g) = 14.0 grams | 1mol N2
28 grams
| 2 mol N2O3
2 mol N2
* moles N2O3(g) = 96.0 grams| 1 mol O2 | 2 mol N2O3
32 grams 3 mol O2
| 76 grams N2O3 | = 38.0 grams N2O3
1 mol N2O3
| 76 grams N2O3 | = 152 grams N2O3
1 mol N2O3
b) In theory, how many grams of the limiting reagent will react? * all 14.0 grams
c) How many grams of the other reactant will react? *24.0 g
How many grams of the other reactant are left in the reaction vessel? * 72.0 g
When asked to find the about “left over”, think about flipping the question: Okay, since I now know
* 14.0 grams of N2 will react, how many grams of the * dioxygen will react with those * 14.0 g N2?
Once you know how much * O2 is consumed, subtract that from the total amount of * O2
what is left over…
to find
*grams O2 react = 14.0 g of N2 | 1 mol N 2 | 3 mol O2 | 32 grams O2| = 24.0 grams O2 react,
28 grams
2 mol N2 1 mol O2
leaving 72g of the original
96.0 grams unreacted....
d) How many grams of dinitrogen trioxide (GFM = 76 g) are produced? * 38.0
* grams N2O3 = 14.0 grams N2 |
1 mol N2 | 2 mol N2O3
28 grams 2 mol N2
grams
| 76 grams N2O3 | = 38.0 grams
1 mol N2O3
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9) Given:
6 I2O5 + 20 BrF3  12 IF5 + 15 O2 + 10 Br2
a) Calculate the maximum mass, in grams of IF5 would be produced using 44.01 grams
of I2O5 and 107.0 grams of BrF3.
* g IF5 = 44.01 g I2O5 | 1 mol I O | 12 mol IF | 222 g IF | = 58.50 grams IF5
2
334 g I O
2
5
5
5
5
6 mol I O 1 mol IF
2
5
5
* g IF5 = 107.0 g BrF3 | 1 mol BrF3 | 12 mol IF | 222 g IF | = 104.0 grams IF5
5
5
137g BrF3 20 mol BrF3 1 mol IF
5
b) What is the excess reagent? * BrF3
grams left over
10) Given: 2 N2(g) + 3 O2(g) 
2 N2O3(g)
Calculate the approximate mass, in grams of the excess reagent remaining in the system
when 75.00 grams of N2(g) are allowed to react with 88.00 grams of O2(g) (hint: find the
limiting reagent and assume it is consumed completely….)
*g N2O3 = 75.00 grams N2 = | 1 mol N2| 2 mol N2O3| 76 grams N2O3 | = 203.6g
28 grams 2 mol N2
1 mol N2O3
*g N2O3 = 88.00 grams O2 = | 1 mol O2| 2 mol N2O3| 76 grams N2O3 | = 139.3 g
32 grams 3 mol N2
1 mol N2O3
*Thus since all 88.00 grams of O2 end up in the 139.3 grams of N2O3 we know that 51.30 grams of
dinitrogen are consumed. This leaves 23.70 grams of dinitrogen gas unreacted.
298
REFLECT: Given all of the practice and homework, the smallest reactant mass is not always the
limiting reagent. So, you really need to do the stoichiometric calculations. What issue(s)
do you think most affect the determination of the limiting reagent?
* the mole ratio (coefficients) coupled to mass of reactants and/or molar masses ….
VII) Percent Yield = actual yield
x 100
theoretical yield
A) If you have ever ended up with a percent error other than 0%, you understand that chemical
reactions can go wrong. There is a difference between the amount you produced and the amount you
should have (theoretically) produced.
1) Procedure:
a) determine the limiting reagent (sometimes it is given ...& sometimes you need to determine it)
b) determine the theoretical yield assuming a perfect reaction using the limiting reagent
value ...but, realize that the theoretical answer of the calculation, is different than the
experimental value (amount of the of the product), given in the problem.
c) determine the actual yield. Often, this is given in the problem.
d) calculate the % yield, using the above equation and two yield values.
2) Practice: (answers are on the next page)
When 84.8 grams of iron (III) oxide (M = 160 g) react with an excess of carbon
monoxide, 52.1 grams of iron (M = 56 g) were experimentally produced.
(NB the CO is in excess, thus Fe2O3 is the limiting reagent, automatically)
Fe2O3 + 3 CO  2 Fe + 3 CO2
a) Calculate the theoretical yield (...just do the stoich problem...)
b) Calculate the % yield
3) When 50.0 g of silicon dioxide (M = 60 g) are heated with an excess of carbon, a student
found that 27.9 g of silicon carbide (M = 40 g) are produced.
SiO2 + 3 C  SiC + 2 CO
What is the percent yield of this reaction?
You need the actual yield (from the problem)
& the theoretical yield from the stoich (math)
299
4) The following reaction has a per cent yield of 96.8%. How many grams of
calcium sulfate (M = 136 g) are formed when 53.4 grams of sulfur dioxide (M = 64 g) reacts with
an excess of calcium carbonate and dioxygen gas?
2 CaCO3(s) + 2 SO2(g) + O2(g)  2 CaSO4(s) + 2 CO2(g)
Answers for percent yield
2a g Fe = 84.8 gFe2O3| 1 mole Fe2O3| 2 mol Fe
| 56 g Fe| = 14.8 g
160 grams
1 mol Fe2O3 1 mol Fe
b) % yield = actual yield
x 100
theoretical yield
% yield = 52.1 x 100
59.4
87.7% yield
3) First calculate the theoretical grams, using silicon dioxide as the limiting reagent.
g SiC = 50.0 g SiO2 | 1mol SiO2| 1mol | 40 grams|
60 grams 1 mol 1 mol SiC
= 33.3 g SiC
thus:
% yield = actual yield
x 100
theoretical yield
% yield = 27.9 x 100
33.3
83.8% yield
4) g CaSO4 = 53.4 g SO2 | 1mol | 2 mol | 136 g CaSO 4| = 113 grams CaSO4 produced (theoretically)
64 g
2 mol
1 mol
With a % yield of 96.8%, or (0.968)(113g) = only 109 grams are produced experimentally (actually)
300
NAME ____________________________________
MORE AND MORE STOICHIOMETRY
(This is Purely Honors Chemistry)
DIRECTIONS: The following problems are based on those found on the OSU website cited in your note
packets. Each of these includes multiple reaction equations. Each of the reaction equations are BALANCED
already. Here's a hint ... When selecting which substances to reference, be sure to select substances common
to each equation. Be sure to link up a reactant and a product in each equation. I have provided some
information, and then BOLDFACED (and underlined) the best substances to use for your first solo problem.
Now, complete each of the following questions. You are in charge... You will need to get creative ... if you
wish to design mole maps, then do so.
Here is the idea:
Example:
Potassium perchlorate, KClO4, may be prepared by reacting KOH with Cl2 and the following series of chemical
reactions. What mass, in grams, of KClO4 (GFM = 138 g) can be prepared using 65.0 g of KOH (GFM = 56 g) in
an excess of Cl2 ?
Start
2 KOH
+ Cl2
KCl + KClO + H2O
3 KClO
2 KCl + KClO3
4 KClO3
3 KClO4 + KCl
The circled formulae are those that
link one step of the synthesis to the
next step…Thus the coefficients of
these substances which are common
to steps (or link one to the other)
become important conversion factors.
Finish
Now look at my solution: desired = given … goalposts drop your units….
I begin with the only measurement given in the problem. I convert to moles of the given and then
use the balanced equations, especially the coefficients of the those substances required to move towards
my goal
g KClO4 = 65.0 g KOH | 1 mol KOH | 1 mol KClO | 1 mol KClO3 | 3 mol KClO4 | 138 g KClO4
56 grams
2 mol KOH
3 mol KClO
4 mol KClO3 1 mol KClO4
*******************************************************
1) The following series of chemical reactions can be used to purify iodine from the saltwater (brine) found in
oil fields, using the NaI in the brine. Calculate the number of moles of I2 produced when 8.50 moles of
AgNO3 are consumed completely.
NaI + AgNO3 → AgI + NaNO3
2 AgI + Fe → FeI2 + 2 Ag
2 FeI2 + 3 Cl2 → 2 FeCl3 + 2 I2
mol I2 = 8.50 mol AgNO3 |
|
mol AgNO3
301
2) Potassium perchlorate (KClO4) may be prepared from KOH and Cl2 by the following series of reactions.
Calculate the moles of KClO4 can be prepared from 8.00 moles of KOH.
2 KOH + Cl2 → KCl + KClO + H2O
3 KClO → 2 KCl + KClO3
4 KClO3 → 3 KClO4 + KCl
3) Sodium chlorate, NaClO3, can be produced by the following series of reactions. How many grams of
sodium chlorate can be produced from 38.5 grams of HCl using an excess of all other reactants?
2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2
6 Cl2 + 6 Ca(OH)2 → Ca(ClO3)2 + 5 CaCl2 + 6 H2O
Ca(ClO3)2 + Na2SO4 → CaSO4 + 2 NaClO3
4) Nitric acid (HNO3) can be prepared by the following series of reactions. How many grams of nitric acid can
be produced from 42.5 grams of NH3?
4 NH3 + 5 O2 → 4 NO + 6 H2O
2 NO + O2 → 2 NO2
3 NO2 + H2O → 2 HNO3 + NO
302
5) Dichlorodifluoromethane (CCl2F2), a refrigerant, can be prepared by the following reactions. Calculate the
volume, in milliliters, of Cl2 gas, at STP, required to produce 1,500.0 grams of CCl2F2
CH4 + 4 Cl2 → CCl4 + 4 HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
6) Dichlorodifluoromethane, yes, that refrigerant, can be prepared by the following two reactions. Calculate the
mass in kilograms of CCl2F2 (M = 120 g) that can be produced from the complete reactions of 60.0 grams of
Cl2 (M =70) in an excess of methane
CH4 + 4 Cl2
CCl4 + 2 HF
CCl4 + 4 HCl
CCl2F2 + 2 HCl
Answers:
1) The following series of chemical reactions can be used to purify iodine from the saltwater (brine) found in oil fields, using the NaI
in the brine. Calculate the number of moles of I 2 produced when 8.50 moles of AgNO 3 are consumed completely.
NaI + AgNO3 → AgI + NaNO3
2 AgI + Fe → FeI2 + 2 Ag
2 FeI2 + 3 Cl2 → 2 FeCl3 + 2 I2
mol I2 = 8.50 mol AgNO3 |
1 molAgI
| 1 mol FeI2| 2 mol I2| = 4.25 mol I2
1 mol AgNO3 2 mol AgI 2 mol FeI2
2) Potassium perchlorate (KClO4) may be prepared from KOH and Cl2 by the following series of reactions.Calculate the moles of
KClO4 can be prepared from 8.00 moles of KOH.
2 KOH + Cl2 → KCl + KClO + H2O
3 KClO → 2 KCl + KClO3
4 KClO3 → 3 KClO4 + KCl
mol KClO4 = 8.00 mol KOH | 1 mol KClO | 1 mol KClO3 | 3 mol KClO4 | = 1.00 mol KClO4
2 mol KOH
3 mol KClO
4 mol KClO 3
303
3) Sodium chlorate, NaClO3, can be produced by the following series of reactions. How many grams of sodium chlorate can
be produced from 38.5 grams of HCl using an excess of all other reactants?
2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2
6 Cl2 + 6 Ca(OH)2 → Ca(ClO3)2 + 5 CaCl2 + 6 H2O
Ca(ClO3)2 + Na2SO4 → CaSO4 + 2 NaClO3
gNaClO3= 38.5 g HCl | 1 mol HCl | 5 mol Cl2 | 1 mol Ca(ClO3)2 | 2 mol NaClO3 |106 g NaClO3|= 11.8 grams
36 HCl
16 mol HCl 6 mol Cl2
1 mol Ca(ClO3)2 1 mol NaClO3
4) Nitric acid (HNO3) can be prepared by the following series of reactions. How many grams of nitric acid can be produced from
42.5 grams of NH3?
4 NH3 + 5 O2 → 4 NO + 6 H2O
2 NO + O2 → 2 NO2
3 NO2 + H2O → 2 HNO3 + NO
g HNO3 = 42.5 gNH3 | 1 mol NH3| 4 mol NO | 2 mol NO2| 2 mol HNO3 | 63 g HNO3| = 105 g HNO3
17 g NH3 4 mol NH3 2 mol NO 3 mol NO2 1 mol HNO3
5) Dichlorodifluoromethane (CCl2F2), a refrigerant, can be prepared by the following reactions. Calculate the volume in liters,
of Cl2 gas, at STP, required to produce 1,500.0 grams of CCl 2F2
CH4 + 4 Cl2 → CCl4 + 4 HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
L Cl2 = 1,500.0 g CCl2F2 | 1 mol CCl2F2 | 1 mol CCl4 | 4 mol Cl2 | 22.4 Liters |1000 mL | = 1.1200 x 106 mL
120 grams
1 mol CCl2F2 1 mol CCl4 1 mol Cl2 1 L
6) Dichlorodifluoromethane, yes, that refrigerant, can be prepared by the following two reactions. Calculate the mass in kilograms
of CCl2F2 (M = 120 g) that can be produced from the complete reactions of 60.0 grams of Cl 2 (M =70) in an excess of
methane
CH4 + 4 Cl2
CCl4 + 4 HCl
CCl4 + 2 HF
kg CCl2F2 =
CCl2F2 + 2 HCl
60.0 g Cl2 | 1 mole Cl2 | 1 mol CCl4 | 1 mol CCl2F2 | 120 grams | 1 kg
| = 0.0257 kg
70 grams
4 mol Cl2
1 mol CCl4
1 mol CCl2F2 1,000 g
304
NAME __________________________________
ADVANCED, ADVANCED STOICH
DIRECTIONS: This is designed to expose you to a slightly more complicated types of stoichiometric
calculations, using dimensional analysis and other concepts from the course. Exemplars are given when the
concept appears nowhere else in your notes.
What If I’m Given a Density of Solution, a Percent Composition and Asked for Molarity? (Independent
Packet on Molarity)
Strategy: This is a molarity problem, so you will need to use M = mol/Liters
You should convert the g/mL density value to g/L
Assume 1 Liter of solution, so multiply the density by 1,000 to get the total solution mass.
Multiply the total solution mass by the percent composition to get the mass of solute
Convert the mass of solute to moles
Since you have assumed 1 Liter of solution, the mole value of solute IS the molarity.
What If I’m Given the Per Cent Compositions of the Elements in a Compound? (Empirical Formula &
Lab)
Strategy: This is an Empirical Formula problem. So:
Use the percentages as masses.
Convert the masses to moles
Divide each mole value by the smallest of all the values
Use the “extra” step, if you need to get whole numbers.
What If I’m Given a Per Cent Composition of a Mixture? (Independent Packet on Percent Composition)
Strategy: Using the percent composition determine how many grams of the total mixture are due
to the desired substance: (% composition)(total mass of mixture)
Use the mass as the “given” for any stoichiometric calculations.
Example: Tetraphosphorus, P4, (M = 124 g) is prepared by the following reaction. What mass, in grams, of
tetraphosphorus can be prepared from the complete reaction of the calcium phosphate from a
1,000. gram mixture that is 86.72% Ca5(PO4)3F (M = 504 g) in an excess of SiO2 and C?
4 Ca5(PO4)3F + 18 SiO2 + 30 C
3 P4 + 2 CaF2 + 18 CaSiO3 + 30 CO
Step1: Find how much of the mixture is reactant (%)(total mass) = mass of reactant
(0.8672)(1,000.) = 867.2 grams of Ca5(PO4)3F
Step 2: Use the answer to step 1 as the given and run the stoichiometry
* grams P4 = 867.2 grams of Ca (PO ) F | 1 mol Ca (PO ) F | 3 mol P
504 g Ca (PO ) F 4 mol Ca (PO ) F
5
4 3
5
5
4 3
4 3
4
5
4 3
| 124 g |
1 mol P4
305
What If I’m Given the Masses of Two (or more) Reactants? (Notes: Limiting Reagent)
Strategy: This is a limiting reagent problem. Do all of the necessary stoichiometric calculations.
The reactant
What If I’m Asked for the Mass of Reactant that is Leftover? (Notes: Limiting Reagent)
Strategy: This is a form of limiting reagent problem.
Find the limiting reagent. Then flip the question to ask: How much of the other
reactant gets consumed assuming the total mass of the limiting reagent is consumed?
Then, use: Mass Leftover = Original mass – Consumed mass
What If I’m Asked for a Percent Yield? (Notes: Limiting Reagent)
Strategy: Determine the limiting reagent. Use that to predict the maximum (theoretical) amount of
product that could be produced. Use that value and the actual yield to determine the percent
yield. This is just another “percentage” such as percent error …or even composition:
% = Part you want to know about x 100
OR
Percent Yield = actual yield
x 100
Whole
theoretical yield
What If I’m Given A Series Of (Multiple) Reaction Equations ? (Notes: More and More Stoich)
Strategy: Identify the key reactants and products to get to the final product. These will be reactants
and products common to the reaction equations.
Use the coefficients of the balanced equations of these reactants and products as conversion
factors in your stoichiometric calculations.
1) Potassium perchlorate, KClO4, may be prepared by reacting KOH with Cl2 and the following series of chemical
reactions. What mass, in grams, of KClO4 (GFM = 138 g) can be prepared using 65.0 g of KOH (GFM = 56 g) in
an excess of Cl2 ?
Start
2 KOH
+ Cl2
KCl + KClO + H2O
3 KClO
2 KCl + KClO3
4 KClO3
3 KClO4 + KCl
The circled formulae are those that
link one step of the synthesis to the
next step…Thus the coefficients of
these substances which are common
to steps (or link one to the other)
become important conversion factors.
Finish
g KClO4 = 65.0 g KOH | 1 mol KOH | 1 mol KClO | 1 mol KClO3 | 3 mol KClO4 | 138 g KClO4
56 grams
2 mol KOH
3 mol KClO
4 mol KClO3 1 mol KClO4
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NAME _____________________________________ MOLE THEORY AND STOICHIOMETRY REVIEW
DIRECTIONS: Hey, returning from any sort of vacation can be tough on the ol' chemistry skills. I often say if you
don't use your chemistry... you lose it ...So, this is just a review for you. Answers are on the last page.
Questions 1-15 deal with the BASICS. (one-step mole conversions, calculating GFM [mole mass], mole-mole
conversions, Avogadro's Hypothesis, and basic balancing by inspection, identification of an empirical formula)
1) Calculate one mole mass for the compound butanoic acid CH3CH2CH2COOH
a) approximatetly 136 grams
c) approximately 91 grams
b) approximately 102 gram
d) approximately 88 grams
2) The formula mass for the compound Mg3(PO4)2 is closest to:
(Note: as a point of review ... the formula mass is just the mass of a single formula unit
of the compound) while a gram formula mass is just the mass of 1 mol of formula units
a) 155 µ
c) 429 µ
b) 262 µ
d) 1,021 µ
3) Calculate the number of grams equivalent to 0.0200 mol of silver.
a) 2.16 grams
b) 3.91 grams
c) 6.10 grams
d) 7.78 grams
4) Which quantity represents 117 grams of LiF?
a)
b)
c)
d)
5) Given:
1.75 moles
2.50 moles
4.50 moles
5.75 moles
2 C2H6(g) + 7 O2(g) 
4 CO2(g) + 6 H2O(l)
How many moles of oxygen are required to combust completely 3.500 moles of C2H6(g)?
a)
b)
c)
d)
17.02 moles
11.82 moles
8.003 moles
12.25 moles
6) At STP, 4.0 liters of H2(g), contain the same number of molecules as
a) 1.5 liters of N2(g)
b) 4.0 liters of N2(g)
c) 6.5 liters of N2(g)
d) 8.0 liters of N2(g)
307
7) Given the balanced equation:
2 C6H14(g) + 19 O2(g)  12 CO2(g) + 14 H2O(l)
A chemist produced 10.0 moles of carbon dioxide. How many moles of water did she also produce?
a) 14.5 moles
b) 9.06 moles
c) 11.7 moles
d) 13.0 moles
8) When comparing 2.0 liters of H2(g) and 2.0 liters of CH4(g) each at 300 K and 1.75 atm, which of the
following is accurate?
a) The samples have equal masses
b) The samples have equal densities
c) The samples have an equal number of atoms
d) The samples have an equal number of molecules
9) Which of the following represents the greatest mass?
a) 1 mol of Cl atoms
1 molecule Cl2
b) 1 Cl atom
c) 1 mol of Cl2 molecules
d)
10) When correctly balanced using the simplest whole number ratio, the coefficient for the chromium solid is:
____ Al(s) + ____ Cr2O3(s)  _____ Cr(s) + ____ Al2O3(s)
a) 1
b) 2
c) 6
d) 4
11) When correctly balanced using the simplest whole number ratio, the coefficient for KNO3 is:
_____Pb(NO3)2
a) 1
b) 2
+ _____ K3PO4 
c) 6
____ Pb3(PO4)2
+ _____KNO3
d) 4
12) When correctly balanced using the simplest whole number ratio, the ratio between
oxygen (dioxygen) and carbon dioxide is:
_____ C8H14
+ ____ O2
 ____ CO2 + ____ H2O
The mole ratio is: __________ to ____________
13) Calculate the per cent water of hydration of: Na2SO4 10 H2O
308
14) When balanced using the simplest whole-number ratio the coefficient for carbon dioxide would be:
____ C3H5N3O9(l)  ____ N2(g) + ____ CO2(g) + ____ H2O(g) + _____ O2(g)
15) When balanced using the simplest whole-number ratio the coefficient for water would be:
____ Rb + _____H2O  ____ H2 + _____ RbOH
a) 1
b) 2
c) 5
d) 6
Questions 16 - 21 deal with basic issues, but connect up with our advanced Honors chemistry work (advanced
stoichiometry, limiting reagents, determination of empirical formula from percent composition)
16) The reaction: 4 KO2(s) + 2 H2O(g) + 4 CO2(g)  4 KHCO3 + 3 O2(g)
is used in jets to generate
emergency oxygen gas, in case of cabin decompression. If you wished to generate 0.250 Liters of O2(g),
at STP, how many grams of the superoxide, KO2 would need to be consumed (reacted)?
Questions 17 - 20 are "themed" or a scaffolded question. Use the reading, the reaction equations and your grasp
of chemistry to complete each question.
The Haber process was designed to produce ammonia (NH3) by using nitrogen gas (dinitrogen N2(g)) from the
atmosphere. Once the ammonia is produced, the ammonia can be converted into fertilizers and explosives.
When the ammonia is heated with a platinum/rhodium catalyst, the ammonia is converted into nitrogen
monoxide (NO) and water. Reacting the nitrogen monoxide with even more dioxygen gas, will produce
nitrogen dioxide, which in turn reacts with water and dioxygen ultimately to produce an aqueous solution
known as nitric acid. Nitric acid is a valuable industrial mixture used as a degreaser, reagent for fertilizers and a
powerful oxidizing agent.
The reaction equations for the production of nitric acid are:
4 NH3 + 5 O2  4 NO + 6 H2O
2 NO + O2 → 2 NO2
4 NO2 + O2 + 2 H2O → 4 Nitric Acid
17) What is the percent composition of nitrogen in a molecule of ammonia?
_________% nitrogen
309
18) Essentially nitric acid is a mixture of water and a rather reactive compound made of hydrogen, nitrogen
and oxygen. That compound is 1.59% hydrogen, 22.2 % nitrogen, and 76.2% oxygen by mass. What is
the empirical formula of the compound?
______________ans.
19) Calculate the maximum number of grams of nitric acid produced when 185.0 grams of ammonia are
reacted completely
____________ grams of nitric acid
20) _______ With how many significant figures should your answer to #19 be recorded?
21) Given: 4 NH3 + 5 O2  4 NO + 6 H2O Assume 20.0 grams of ammonia are reacted with 55.0 g of O2(g)
a) Identify the limiting reagent of this reaction. ________
b) Calculate the maximum number of grams of NO that can be produced, with the above quantities.
_____________ grams NO
310
Answers:
1) d
4 mol C (12 grams) + 8 mol H (1 gram) + 2 mol (16 gram) =
2) b
3 mol Mg (24 µ ) + 2 mol P (31 µ ) + 8 mol O (16 µ ) =
3) a
moles = 0.0200 mol Ag | 108 grams |
1 mol
4) c
moles LiF = 117 grams LiF | 1 mol |
26 grams
5) d
moles O2 = 3.500 mol C2H6(g) | 7 mol O2 |
2 mol C2H6(g)
6) b Gas samples have the same number of molecules, with same volume, same temperature, same pressure...
7) c
moles of water = 10.0 moles CO2 | 14 moles water |
12 moles CO2
8) d Avogadro's Hypothesis.... remember ... masses are very rarely equal....
9) c
10) b
11) c
12) 23 to 16 or 23:16
13) 55.9%
14) 12
15) b
16) grams of KO2 = 0.250 L O2 | 1 mol oxygen | 4 moles KO2 | 71 grams |
22.4 Liters
3 moles O2
1 mol KO2
= 1.06 grams Answer
17) approx. 82.4 % nitrogen (82.35% is great as well)
% = Part x 100
Whole
18 HNO3
or % = 14 x 100
17
Find Moles: H: 1.59/1 = 1.59
Find Ratio: 1.59/1.59 = 1
N: 22.2/14 = 1.59
1.59/1.59 = 1
O 76.2/16 = 4.76
4.76 / 1.59 = 2.99 = 3
311
19) grams HNO3 = 185.0 grams NH3 | 1 mol NH3 | 4 mol NO | 2 mol NO2 | 4 mol HNO3 | 63 grams |
17 grams 4 mol NH3 2 mol NO 4 mol NO2
1 mol HNO3
= 685.6 grams Answer
20) 4 sig figs, due to the 4 sig figs in the GIVEN value.
21) You can do this different ways .... You can use the multiplier system to find the smaller value (this would
be the limiting reagent) ... or you can run the stoichiometry for both reactants and the one with the smaller
production of product is the limiting reagent...
Multiplier: convert to moles and divide by the coefficient of the balanced equation…
moles NH3 = 20.0 grams | 1 mol | = 1.18 mol
17 grams
1.18 NH3
4
vs.
moles O2 = 55.0 grams | 1 mol| = 1.72 mol
32 grams
1.72 grams O2
5
(this is the coefficient of oxygen from the bal. equ)
(4 is the coefficient of ammonia from the bal. equ)
multiplier = 0.295
and
0.344
since the multiplier for ammonia is the lesser value, then the
ammonia is the limiting reagent.
Since you know the ammonia is the limiting reagent, use the 20.0 grams as your GIVEN value,
and calculate the grams of NO produced.
b) grams NO = 20.0 grams NH3 | 1 mol NH3 | 4 mol NO | 30 grams NO |
17 grams
4 mol NH3
1 mol NO
= 35.3 grams NO produced. Answer
If you wished you could have just done the second stoichiometry and noted that the mass of NO produced
is greater than with ammonia ... hence the oxygen could not be the limiting reagent....
grams NO = 55.0 grams O2 | 1 mol O2 | 4 mol NO | 30 grams NO |
32 grams O2
5 mol O2
1 mol NO
= 41.3 grams NO ... which is more than produced with ammonia (confirming ammonia is the
limiting reagent and proving that oxygen is not the limiting reagent).
Compare: 35.3 grams of NO produced using the 20.0 g of ammonia VS.
41.3 grams of NO produced using the 55.0 g of oxygen
312