LIMITING REAGENT PROBLEMS
The first step in solving a limiting reagent problem is being able to recognize that
you have a limiting reagent problem.
Suppose you were given the following problem:
A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the
reaction:
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
What is the theoretical yield of MgCl2?
Is this a limiting reagent problem? One way to find out is to write down what is
known about any component of the reaction below that component:
Mg(OH)2
50.6 g
+
2 HCl
45.0 g
MgCl2
?g
+
2 H2O
Notice how quantities of both reactants are known. Which one will be used up
first? You can't tell, nor should you jump to any conclusions. Just because it looks like
there is less Mg(OH)2 present does not automatically mean it will be used up before all of
the HCl is consumed. This is a limiting reagent problem.
There are numerous ways to solve a limiting reagent problem. I will present 3 different
methods. You choose the method that you feel most comfortable with.
Method 1
In order to find out which reactant is the limiting reagent, you have to compare
them to each other. This comparison must be done in moles; therefore, the next step will
be to convert each of the grams of reactants to moles using their molar mass:
50.6 g Mg(OH) 2 x
1 mol Mg(OH) 2
 0.868 mol Mg(OH) 2
58.3 g Mg(OH) 2
45.0 g HCl x
1 mol HCl
 1.23 mol HCl
36.5 g HCl
Again, you should not jump any conclusions about which reactant is the limiting reagent.
Just because there are fewer moles of magnesium hydroxide does not mean it is the
limiting reagent. Arbitrarily pick one of these reactants and calculate how many moles of
the other reactant is needed to completely use up the reactant picked. In this case,
magnesium hydroxide is arbitrarily chosen:
0.868 mol Mg(OH) 2 x
2 mol HCl needed
 1.74 mol HCl needed
1 mol Mg(OH) 2
Compare the moles HCl needed to the actual moles HCl available. In this case,
1.74 mole of HCl is needed and 1.23 mole HCl is available--that's not enough. So, even
though it appears that there are more moles of HCl than Mg(OH)2, the HCl is the limiting
reagent. The HCl will be run out before the magnesium hydroxide and thereby limit the
amount of product formed. For this reason, use the moles of HCl to calculate the
theoretical yield of magnesium chloride:
1.23 mol HCl x
1 mol MgCl 2 95.3 g MgCl 2
x
 58.6 g MgCl 2
2 mol HCl
1 mol MgCl 2
The theoretical yield is the maximum amount of product that can be produced (in
an ideal world). In the "real" world it is difficult to produce the amount obtained for the
theoretical yield. A percent yield is often used to show how close to ideality one has
obtained in a chemical synthesis. Suppose in the reaction discussed a chemist actually
obtained 55.4 g of MgCl2. This is called the actual yield and would be given to you in the
problem. To calculate the percent yield:
% yield 
55.4 actual g MgCl 2
x 100  94.5 %
58.6 theoretic al g MgCl 2
Method 2
An alternative method involves comparing the theoretical ratio of the reactants to
the actual ratio of reactants available. First, calculate the moles of each reactant using
their molar mass:
50.6 g Mg(OH) 2 x
1 mol Mg(OH) 2
 0.868 mol Mg(OH) 2 available
58.3 g Mg(OH) 2
45.0 g HCl x
1 mol HCl
 1.23 mol HCl available
36.5 g HCl
Consider the balanced reaction:
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
From the balanced equation, the theoretical mole ratio is:
2 moles of HCl needed
1 mol Mg(OH) 2
Let’s see what the actual mole ratio is based on actual the amounts of reactants
present:
1.23 mol HCl
1.42 mol HCl present

0.868 mol Mg(OH) 2
1 mol Mg(OH) 2
Comparing these two ratios, one can easily see that there is not enough HCl, so HCl must
be the limiting reagent:
1.42 mol HCl present
2 mol HCl needed
compared to
1 mol Mg(OH) 2
1 mol Mg(OH) 2
Since HCl is the limiting reagent, use the moles of HCl available to calculate the
theoretical yield of MgCl2:
1.23 mol HCl x
1 mol MgCl 2 95.3 g MgCl 2
x
 58.6 g MgCl 2
2 mol HCl
1 mol MgCl 2
Method 3
A third method is to calculate the theoretical yield of product produced by each
reactant and chose the lesser amount:
50.6 g Mg(OH) 2 x
1 mol Mg(OH) 2
1 mol MgCl 2
95.3 g MgCl 2
x
x
 82.7 g MgCl 2
58.3 g Mg(OH) 2 1 mol Mg(OH) 2 1 mol MgCl 2
1 mol HCl 1 mol MgCl 2 95.3 g MgCl 2
x
x
 58.6 g MgCl 2
36.5 g HCl
2 mol HCl
1 mol MgCl 2
Since HCl produced less product, HCl is the limiting reagent and 58.6 g MgCl2 is the
theoretical yield.
45.0 g HCl x
Stoichiometry Problem Set
1. Calculate the molar mass of the following compounds to two decimal places.
a.
b.
c.
d.
e.
AlCl3
Na2SO4
Cr2(CO3)3
C7H5N3O6
CuSO4∙5H2O
2. Given 115 grams of Na2SO4, find:
a.
b.
c.
d.
e.
the number of moles of sodium sulfate.
the number of moles of sodium ions
the number of moles of sulfate ions
the number of moles of sulfur
the number of moles of oxygen
3. Given 3.5 moles of K2CO3, find:
a.
b.
c.
d.
e.
the number of grams of potassium carbonate
the number of moles of potassium ions
the number of moles of carbonate ions
the number of potassium ions
the number of carbon atoms
4. Balance the following reactions by inspection:
a.
S8
+
F2 → SF4
b.
CH4
+
Cl2 → CCl4
c.
C5H12
d.
Ba(NO3)2
e.
CaCl2
→
+ O2
+
+ HCl
CO2
→
Na2CO3
+
K3PO4
+
→
H2O
BaCO3
+
NaNO3
Ca3(PO4)2
+
KCl
5. If 12.5 grams of propane, C3H8, react with excess oxygen according to the following
unbalanced reaction:
C3H8
+
O2
→
CO2
+
H2O
Calculate the following:
a.
b.
c.
d.
e.
the moles of oxygen gas needed
the moles of carbon dioxide formed
the grams of carbon dioxide formed
the percent yield, if 35.2 grams of carbon dioxide are actually isolated
the molecules of carbon dioxide formed
6. If 5.39 x 1024 copper atoms react with excess nitric acid according to the reaction:
Cu(s)
+ 4 HNO3(aq)
→
Cu(NO3)2(aq) +
2NO2(g)
+
2 H2O(l)
Calculate the following:
a.
b.
c.
d.
e.
the number of grams of copper atoms
the number of moles of nitric acid needed
the number of grams of copper(II) nitrate formed
the number of molecules of nitrogen dioxide formed
the number of moles of water formed
7. If 6.20 grams of ethane, C2H6, reacts with 92.0 grams bromine liquid to form
hexabromoethane, C2Br6 and hydrogen bromide gas according to the following reaction:
C2H6 (g)
a.
b.
c.
d.
e.
+ 6 Br2 (l) → C2Br6 (l)
+ 6 HBr (g)
Find the limiting reactant
Calculate the moles of HBr formed
Calculate the theoretical yield of C2Br6
If the actual yield of hexabromoethane is 45.9 grams, what is the percent yield?
Calculate the grams of reactant left over in excess
8. If 14.7 grams of copper reacts with 450.0 mL of 2.50 M HNO3 according to the reaction:
Cu(s)
+ 4 HNO3(aq)
→
Cu(NO3)2(aq)
+
2NO2(g)
+ 2 H2O(l)
Calculate the following:
a. the grams of copper(II) nitrate formed
b. the molarity of the the copper(II) nitrate in solution (assume volume of Cu is
negligible)
c. the molecules of nitrogen dioxide formed
d. the moles of excess reactant
e. If 39.5 grams of copper(II) nitrate is isolated, what is the percent yield?
Answers
Most stoichiometry problems follow the strategy:
Quantity A → mol A → mol B→ Quantity B
You will be using portions of this strategy or the entire strategy to complete this problem set.
1. Sum the atomic weights (AW) of each of the elements in the compound according to the
number of each element present. The molar mass is defined as the number of grams of substance
per mole of substance. It has the same mass in g/mol as one formula unit in amu.
a. AlCl3
Molar mass = AW Al + 3(AW Cl) = 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol
b. Na2SO4
Molar mass = 2(AW Na) + AW S + 4(AW O) = 2(22.99 g/mol) + 32.07 g/mol +
4(16.00 g/mol) = 142.05 g/mol
c. Cr2(CO3)3
Molar Mass = 2(AW Cr) + 2(AW C) + 6(AW O) = 2(52.00 g/mol) + 3(12.01
g/mol) +
9(16.00 g/mol) = 284.03 g/mol
d. C7H5N3O6
Molar Mass = 7(AW C) + 5(AW H) + 3(AW N) + 6(AW O) = 7(12.01 g/mol) +
5(1.01 g/mol) + 3(14.01 g/mol) + 6(16.00 g/mol) = 227.15 g/mol
e. CuSO4∙5H2O (a hydrate)
Molar Mass = AW Cu + AW S + 9(AW O) + 10(AW H) = 63.54 g/mol + 32.07
g/mol +
10(1.01 g/mol) + 9(16.00 g/mol) = 249.71 g/mol
2. a. Use the molar mass of sodium sulfate as a “conversion factor” to calculate the moles of
sodium sulfate. In terms of the stoichiometry strategy, the molar mass allows you to convert a
Quantity A to mol A or mol B to Quantity B. There are 142.05 grams of Na2SO4 in 1 mole of
Na2SO4. Let the units decide how to set up the problem.
1 mol Na2 SO4
115 g Na2 SO4 x
= 0.810 Na2 SO4
142.05 g Na2 SO4
b. Here we are taking Quantity A to mol A to mol B. The subscript of 2 in the formula
for Na2SO4 indicates there are 2 moles of sodium ions per mole of sodium sulfate. Let
the units help you set up the problem:
115 g Na2 SO4 x
1 mol Na2 SO4
2 mol Na+
x
= 1.62 mol Na+
142.05 g Na2 SO4 1 mol Na2 SO4
c. Again, here we are taking Quantity A to mol A to mol B. In mole of sodium sulfate
there is one mole of sulfate ion. The units help you set up the problem in a similar
fashion as in problem 2a:
2−
1 mol Na2 SO4
1 mol SO4
115 g Na2 SO4 x
x
= 0.810 mol SO2−
4
142.05 g Na2 SO4 1 mol Na2 SO4
d. In one mole of sodium sulfate there is 1 mole of sulfur as indicated by the subscript of
one in the formula. Once again, units help set up the problem for converting Quantity A
to mol A to mol B:
115 g Na2 SO4 x
1 mol Na2 SO4
1 mol S
x
= 0.810 mol S
142.05 g Na2 SO4 1 mol Na2 SO4
e. In mole of sodium sulfate there are four moles of oxygen as indicated by the
subscript of four in the formula. Once again, units help set up the problem for converting
Quantity A to mol A to mol B:
115 g Na2 SO4 x
1 mol Na2 SO4
4 mol O
x
= 3.24 mol O
142.05 g Na2 SO4 1 mol Na2 SO4
3 a. Here we already know mol A and will convert to mol B to Quantity B of the
strategy. We will need the molar mass of potassium carbonate to do the mol B to
Quantity B conversion:
Molar mass = 2(AW K) + AW C + 3(AW O) = 2(39.10 g/mol) + 12.01
g/mol +
3(16.00 g/mol) = 138.21 g/mol
Let the units help you set up the problem. One mole of potassium carbonate has a mass
of 138.21 grams:
3.5 mol K 2 CO3 x
138.21 g K 2 CO3
= 4.8 x 102 g K 2 CO3
1 mol K 2 CO3
b. Convert mol A to mol B. In one mole of potassium carbonate there are two
potassium ions:
2 mol K +
3.5 mol K 2 CO3 x
= 7.0 mol K +
1 mol K 2 CO3
c. Once again, convert mol A to mol B. In one mole of potassium carbonate there is
one mole of carbonate:
3.5 mol K 2 CO3 x
1 mol CO2−
3
= 3.5 mol CO2−
3
1 mol K 2 CO3
d. Here, the number of particles of potassium ions is required. This tips you off that
Avogadro’s number is needed somewhere in the problem. Avogadro’s number is the
number of particles in one mole of substance. There will be 6.02 x 1023 potassium ions in
one mole of potassium ions. Convert mol A to mol B to Quantity B. In this case, mol
potassium carbonate to mol potassium ions to number of potassium ions:
2 mol K +
6.02 x 1023 K + ions
3.5 mol K 2 CO3 x
x
= 4.2 x 1024 K + ions
1 mol K 2 CO3
1 mol K +
e. This problem is similar to 3d. Convert mol A to mol B to Quantity B. In this
case, mol potassium carbonate to mol carbon to number of carbon atoms:
1 mol C
6.02 x 1023 C atoms
3.5 mol K 2 CO3 x
x
= 2.1 x 1024 C atoms
1 mol K 2 CO3
1 mol C
4. When balancing chemical reactions there are a few things to keep in mind. Never
change the subscripts in the formulas found in the reaction. Place coefficients in
front of the substance, never in the middle of the substance. Polyatomic ions may be
balanced as a unit, if they are unchanged on both sides of the reaction. There are a
few tricks one can use to quickly balance chemical reactions. If an element is off by
itself, balance that element last. You can always put any number needed in the end
to balance that element. If there is an odd number of an element on one side the
reaction, and an even number on the other, make the odd side even by multiplying
by two.
a. Balance the sulfur first, as there is an odd number on the right hand side and an
even number on the left hand side. In this case, don’t multiply by 2, rather balance it
with an 8:
S8
+
F2 →
8 SF4
Now the sulfur is balance, but there are 2 F on the left hand side and 32 F on the right hand side.
Balance the F with a coefficient of 16 on the left hand side:
S8
+ 16 F2 →
8 SF4
b. Balance the Cl last, as it found as an element on the left hand side. The carbon is
already balanced, so balance the H. There are 4 H on the left hand side, so place a
coefficient of 4 with the HCl to balance the H:
CH4
+
Cl2 → CCl4
+ 4 HCl
The carbon is still balanced along with the H. There are 2 Cl on the left hand side
and 8 Cl on the right hand side Don’t forget to look at all the products – 4 Cl are
found in CCl4 and 4 Cl are with the 4 HCl. Balance the Cl by placing a coefficient of 4
with the Cl on the left hand side:
CH4
+
4 Cl2 →
CCl4
+ 4 HCl
c. Oxygen is off by itself, so balance it last. There are 5 C on the left hand side and
1 C on the right hand side. Balance the carbon by placing a coefficient of 5 in front of
CO2:
C5H12
→
+ O2
5 CO2
+
H2O
Next balance the H. There are 12 H on the left hand side and 2 H on the right hand
side. Place a coefficient of 6 in front of H2O to balance the H:
C5H12
+
O2
→
5 CO2
+
6 H2O
Lastly, balance the O. There are 2 O on the left hand side and 16 O on the right hand
side. Place a coefficient of 8 in front of O2 to balance the O:
C5H12
+ 8 O2
→
5 CO2
+
6 H2O
d. In this problem, there isn’t any easy place to start. To keep things simpler,
notice that the polyatomic ions can be balanced as the ion, rather than as separate
elements. Barium is balanced, but there are 2 NO3─ ions on the left hand side and 1
NO3─ ion on the right hand side. Place a coefficient of 2 in front of NaNO3 to balance
nitrate:
Ba(NO3)2
+
Na2CO3
→
BaCO3
+
2 NaNO3
This also balances the Na. There are now 2 Na on both sides of the reaction. Also,
CO32- is balanced as well. There is one carbonate in both sides of the reaction.
e. Once again, there isn’t an easy place to start, but you should recognize that
phosphate ions, PO43- can be balanced as the polyatomic ion. Start with balancing
the calcium (arbitrary). There is 1 Ca on the left hand side and 3 Ca on the right
hand side. Place a coefficient of 3 in front of CaCl2 to balance the calcium:
3 CaCl2
+
K3PO4
→
Ca3(PO4)2
+
KCl
From here, continue on sequentially. There are now 6 Cl on the left hand side and 1 Cl on the
right hand side. Balance the Cl by placing a coefficient of 6 in front of KCl to balance the Cl:
3 CaCl2
+
K3PO4
→
Ca3(PO4)2
+
6 KCl
Now there are 6 K on the right hand side and 3 K on the left hand side. Balance the K by placing
a coefficient of 2 in front of K3PO4:
3CaCl2
+
2 K3PO4
→
Ca3(PO4)2
+
6 KCl
This also balances the PO43-. There are now 2 phosphate ions on the left hand side and 2
phosphate ions on the right hand side.
5. Balance the reaction first: C3H8 + 5 O2 → 3 CO2
+ 4 H2O. We will be back
to using the stoichiometry strategy, Quantity A → mol A → mol B → Quantity B.
a. The molar mass of propane, C3H8, will be needed:
Molar mass = 3(AW C) + 8(AW H) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11
g/mol
Use the molar mass to convert grams of propane to mol of propane and the balance reaction to
convert mol of propane to mol of oxygen gas needed. From the strategy, we are using Quantity A
→ mol A → mol B:
1 mol C3 H8
5 mol O2
12.5 g C3 H8 x
x
= 1.42 mol O2
44.11 g C3 H8 1 mol C3 H8
b. From the balanced reaction, 3 moles of carbon dioxide are formed from 1 mole of propane.
Once again, we are converting grams of propane to moles of propane using its molar mass and
then converting moles of propane to moles of carbon dioxide. From the strategy, we are using
Quantity A → mol A → mol B:
1 mol C3 H8
3 mol CO2
12.5 g C3 H8 x
x
= 0.850 mol CO2
44.11 g C3 H8 1 mol C3 H8
c. Here we will use the entire strategy. Convert the grams of propane to moles of propane, then
moles of propane to moles of carbon dioxide and finally moles of carbon dioxide to grams of
carbon dioxide. We will need to calculate the molar mass of carbon dioxide. All other
information has already been previously found in parts a – c of the problem:
Molar mass of CO2 = AW C + 2(AW O) = 12.01 g/mol + 2(16.00 g/mol) = 44.01
g/mol
12.5 g C3 H8 x
1 mol C3 H8
3 mol CO2
44.01 g CO2
x
x
= 37.4 g CO2
44.11 g C3 H8 1 mol C3 H8
1 mol CO2
d. The mass of carbon dioxide calculated in part c of this problem is often referred to as the
theoretical yield. The theoretical yield represents the maximum amount of product that can be
produced in an ideal world with ideal chemists who can collect every single particle of product
formed. Needless to say in the real world this doesn’t happen. As a result one often can calculate
a percent yield. A percent is always a “part/whole” ratio. The actual yield, 35.2 g CO2 represents
the “part” and the theoretical yield, 37.4 g CO2, represents the “whole.” To calculate the percent
yield:
% yield =
actual yield
35.2 g CO2
x 100 =
x 100 = 94.1 % yield
theoretical yield
37.4 g CO2
e. Once again, the entire strategy is used. Since we are asked for the number of
molecules (particles) of CO2, Avogadro’s number will be needed. Convert the mass
of C3H8 to moles of C3H8 to moles of CO2 to molecules of CO2:
23
12.5 g C3 H8 x
1 mol C3 H8
3 mol CO2
6.02 x 10
x
x
44.11 g C3 H8 1 mol C3 H8
1 mol CO2
= 5.1 x 1023 molecules CO2
6. a. Convert the number of copper atoms to moles copper to grams copper:
5.39 x 1024 Cu atoms x
1 mol Cu
63.54 g Cu
x
= 569 g Cu
23
6.02 x 10 Cu atoms
1 mol Cu
b. Convert the number of copper atoms to moles copper to moles of nitric acid:
5.39 x 1024 Cu atoms x
1 mol Cu
4 mol HNO3
x
= 35.81 mol HNO3
6.02 x 1023 Cu atoms
1 mol Cu
c. Convert the number of copper atoms to moles copper to moles copper(II)
nitrate to grams of copper(II) nitrate using the balanced reaction:
5.39 x 1024 Cu atoms x
1 mol Cu
1 mol Cu(NO3 )2 187.56 g Cu(NO3 )2
x
x
6.02 x 1023 Cu atoms
1 mol Cu
1 mol Cu(NO3 )2
= 1.68 x 103 g Cu(NO3)2
d. Since molecules are particles and directly analogous to moles, one can convert
molecules copper directly to molecules nitrogen dioxide using the coefficients in the
balanced reaction:
5.39 x 1024 Cu atoms x
2 NO2 molecules
= 1.08 x 1025 NO2 molecules
1 Cu atom
e. Convert the atoms of copper to moles of copper to moles of water using the
balanced reaction:
5.39 x 1024 Cu atoms x
1 mol Cu
2 mol H2 O
x
= 17.9 mol H2 O
6.02 x 1023 Cu atoms
1 mol Cu
7. A limiting reactant problem is recognizable because quantities of both reactants
are given in the problem. One of these reactants will be used up before the other is
totally consumed.
a. To find the limiting reactant, first find the moles of each reactant so that they
may be compared. Don’t jump to any conclusions about the limiting reactant. The
reactant with a lower mass isn’t always the limiting reactant.
6.20 g C2 H6 x
92.0 g Br2 x
1 mol C2 H6
= 0.206 mol C2 H6
30.08 g C2 H6
1 mol Br2
= 0.576 mol Br2
159.82 g Br2
Once again, don’t jump to any conclusion about the limiting reactant. Even though it
appears that there are fewer moles of ethane, a comparison has not been made to
establish which reactant will be consumed first. Pick one of the reactants and
calculate how many moles of the other reactant is needed. Let’s arbitrarily choose
ethane and calculate the moles of bromine needed using the balanced reaction:
0.206 mol C2 H6 x
6 mol Br2 needed
= 1.24 mol Br2 needed
1 mol C2 H6
Although it appears that there is a greater amount of bromine present, bromine
turns out to be the limiting reactant: 0.618 moles of bromine are needed to react all
0.206 mol of ethane, and only 0.576 mole of bromine is available.
b. Use the moles of limiting reactant and the balanced reaction to calculate the
moles of HBr formed:
6 mol HBr
0.576 mol Br2 x
= 0.576 mol HBr
6 mol Br2
c. Using the moles of limiting reactant and the balanced reaction to calculate the
theoretical yield in grams of C2Br6:
0.576 mol Br2 x
1 mol C2 Br6 503.74 g C2 Br6
x
= 48.4 g C2 Br6
6 mol Br2
1 mol C2 Br6
d. A percent is always a “part/whole” ratio. Use the actual yield as the “part” and the
theoretical yield as the “whole”:
actual yield
45.9 g actual
% yield =
x 100 =
x 100 = 94.9 % yield
theoretical yield
96.7 g theoretical
e. The reactant in excess was ethane. Find the moles of ethane that actually reacted
using the balanced reaction:
1 mol C2 H6 used
0.576 mol Br2 x
= 0.0960 mol C2 H6 used
6 mol Br2
Subtract the mole ethane used from the total moles of ethane available:
0.206 mol C2H6 – 0.0960 mol C2H6 = 0.110 mol C2H6 in excess
Convert the moles of ethane in excess to grams:
0.110 mol C2 H6 x
30.08 g C2 H6
= 3.31g C2 H6 in excess
1 mol C2 H6
8. a. This is a limiting reactant problem because the quantities of both reactants are
given. It is impossible to tell which reactant is used up before the other. A comparison
of both reactants must be made in moles to determine the limiting reactant. Convert each
of the reactant quantities to moles. The atomic weight of copper is used to convert grams
copper to moles copper:
1 mol Cu
14.7 g Cu x
= 0.231 mol Cu
63.54 g Cu
Think of molarity as a conversion factor that allows you to convert moles of solute to
liters of solution or in this case, volume of solution to moles of solute:
1 x 10−3 L 2.50 mol HNO3
450.0 mL soln x
x
= 1.12 mol HNO3
1 mL soln
1 L soln
Choose one reactant and calculate the moles of the other reactant needed. In this case,
mole of copper is arbitrarily chosen and the mole of nitric acid needed is calculated:
0.231 mol Cu x
4 mol HNO3 needed
= 0.924 mol HNO3 needed
1 mol Cu
Compare the moles of nitric acid needed to the moles of nitric acid available: 0.924 mol
of HNO3 are needed and 1.12 mol of HNO3 is available. It is clear that HNO3 is in
excess, thus Cu is the limiting reactant. Use the mole of limiting reactant to calculate the
theoretical yield in grams of copper(II) nitrate formed:
0.231 mol Cu x
1 mol Cu(NO3 )2 187.56 g Cu(NO3 )2
x
= 43.3 g Cu(NO3 )2
1 mol Cu
1 mol Cu(NO3 )2
b. The total volume of the solution is 450.0 mL since we are assuming the volume
copper is not significantly adding to the total volume of the solution. First find the moles
of Cu(NO3)2 formed and divide the moles of solute by the volume in liters:
0.231 mol Cu x
1 mol Cu(NO3 )2
= 0.231 mol Cu(NO3 )2
1 mol Cu
1 x 10−3 L
450.0 mL soln x
= 0.4500 L solution
1 mL soln
0.231 mol Cu(NO3 )2
= 0.531 M Cu(NO3 )2
0.4500 L soln
c. Convert the mole of limiting reactant to mole of nitrogen dioxide to molecules of
nitrogen dioxoide:
2 mol NO2 6.02 x 1)23 molecules NO2
0.231 mol Cu x
1 mol Cu
1 mol NO2
23
= 2.78 x 1O molecules NO2
d. The mole of HNO3 needed and the total mole of HNO3 available has already been
found in part a of the problem. To find the mole of HNO3 in excess, subtract the mole of
HNO3 needed from the total mole of HNO3 available:
1.12 mol HNO3 – 0.924 mol HNO3 = 0.20 mol HNO3 in excess
e. A percent is always a “part/whole” ratio. Use the actual yield as the “part” and the
theoretical yield as the “whole”:
actual yield
39.5 g actual
% yield =
x 100 =
x 100 = 91.2 % yield
theoretical yield
43.3 g theoretical