# stoicheometry - ```U
238.029
Stoichiometry
Worksheets
Stoichiometry
Chemistry worksheets,
Mole Ratios
Stoichiometry: The study of quantitative relationships between the amounts of reactants
used and amounts of products formed by a chemical reaction.
Stoichiometry is based on the law of conservation of mass. Recall that the law states that
matter is neither created nor destroyed in a chemical reaction. In any chemical reaction, the
amount of matter present at the end of the reaction is the same as the amount of matter
present at the beginning. Therefore, the mass of the reactants equals the mass of the products.





3O2(g)


iron
oxygen

4 atoms Fe
3 molecules O2

4 mol Fe
3 mol O2

223.4 g Fe
96.00 g O2
319.4 g reactants
4Fe(s)
2Fe2O3(s)
iron(III) oxide
2 formula units Fe2O3
2 mol Fe2O3
319.4 g Fe2O3
319.4 g products
1) In the following interactions, determine the number of moles of each component of the
reaction.
a)
2PbO2

2PbO

O2
b)
2Fe(OH)3

Fe2O3

3H2O
c)
(NH4)2CO3

2NH3

H2O

CO2
d)
4H2

Fe3O4

3Fe

4H2O
Law of conservation of mass
#𝒈
Molar mass ratio
𝟏𝒎𝒐𝒍
2) conclude the molecular mass relationship of the following Elements
Substance
Molar mass
Substance
Molar mass
Substance
H
N
N2
O
Fe
Mg
O2
C
Cl
O3
S8
Li
pg. 1
7g/mol
Molar mass
Stoichiometry
Chemistry worksheets,
3) conclude the molecular mass relationship of the following compounds
Substance
Molar mass
Substance
Molar mass
Substance
H2O
18g/mol
CO2
(NH4)2CO3
Fe2O3
PbO2
Zn(NO3)2
C3H8
CH4
Fe(OH)3
Molar mass
4) The combustion of propane (C3H8) provides energy for heating homes, cooking food, and
soldering metal parts. Interpret the equation for the combustion of propane in terms of
representative particles, moles, and mass. Show that the law of conservation of mass is
observed.
C3H8 (g) + 5 O2 (g)  3CO2 (g) + 4 H2O(g)
1 molecule C3H8+ 5 molecules O2  3 molecules CO2 + 4 molecules H2O
The coefficients
also indicate the
number of moles.
1 mol C3H8+ 5 mol O2  3 mol CO2 + 4 mol H2O
moles of reactant or product &times;
𝑔𝑟𝑎𝑚𝑠 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑜𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
1 𝑚𝑜𝑙 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑜𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
The coefficients
indicate the number
of molecules.
= grams of reactant or product
The conversion factor we use to calculate mass
1 mol C3H8 &times;
5 mol O2 &times;
44.09 𝑔 𝐶3𝐻8
1 𝑚𝑜𝑙 𝐶3𝐻8
32.00 𝑔 𝑂2
1 𝑚𝑜𝑙 𝑂2
3 mol CO2 &times;
4 mol H2O &times;
= 44.09 g C3H8
Calculate the mass of the reactant C3H8 .
Calculate the mass of the reactant O2
= 160.0 g O2.
44.01 𝑔 𝐶𝑂2
1 𝑚𝑜𝑙 𝐶𝑂2
18.02 𝑔 𝐻2𝑂
1 𝑚𝑜𝑙 𝐻2𝑂
Calculate the mass of the product CO2
= 132.0 g CO2.
Calculate the mass of the product H2O
= 72.08 g H2O
44.09 g C3H8 + 160.0 g O2 = 204.1 g reactants.
pg. 2
Add the masses of the reactants
Chemistry worksheets,
132.0 g CO2 + 72.08 g H2O = 204.1 g products.
204.1 g reactants = 204.1 g products.
pg. 3
Stoichiometry
Add the masses of the products
the law of conservation of mass is observed
Stoichiometry
Chemistry worksheets,
Calculate the mass of the reactants and products in the following reactions and apply the
law of conservation of mass
→
5) 2PbO2
6) 2Fe(OH)3
7) (NH4)2CO3
2PbO
→
→
+
O2
Fe2O3
+
3H2O
2NH3
+
H2 O
+
CO2
Balance the following equations and apply the law of conservation of mass
8) H2
pg. 4
+
O2
→
H2 O
Stoichiometry
Chemistry worksheets,
9) CH4
+
O2 →
CO2
+
H2 O
Mole ratio
A mole ratio is a conversion factor that relates the amounts in moles of any two substances in
a chemical reaction. The numbers in a conversion factor come from the coefficients of the
balanced chemical equation.
The production of ammonia (NH3) from nitrogen and hydrogen gases is an important industrial
reaction called the Haber process, after German chemist Fritz Haber.
N2(g)+3H2(g)→2NH3(g)
The balanced equation can be analyzed in several ways, as shown in Figure below.
The following six mole ratios can be written for the ammonia forming reaction above.
1 mol N2 3 mol H2
1 mol N2
2 mol NH3 2 mol NH3
3 mol H2
,
,
,
,
,
3 mol H2 1 mol N2 2 mol NH3
1 mol N2
3 mol H2
2 mol NH3
Note that the number of mole ratios you can write for a chemical reaction involving a total of
n substances is (n)(n-1). Thus, for reactions involving four and five substances, you can write
12 and 20 moles ratios, respectively.
Four substances: (4)(3) = 12 mole ratios
Five substances: (5)(4) = 20 mole ratios
pg. 5
Chemistry worksheets,
Stoichiometry
10) Determine all possible mole ratios for the following balanced chemical equations.
a) 4Al(s) + 3O2(g) → 2Al2O3(s)
b) 3Fe(s) + 4H2O(l) → Fe3O4(s) + 4H2(g)
c) 2HgO(s) → 2Hg(l) + O2(g)
11) Balance the following equations and determine the possible mole ratios.
a) ZnO(s) + HCl(aq) → ZnCl2(aq) + H2O(l)
b) butane ( C4H10 ) + oxygen → carbon dioxide + water
pg. 6
Stoichiometry
Chemistry worksheets,
Stoichiometric Calculations
Mole to mole conversion
In a mole ratio problem, the given substance, expressed in moles, is written first. The
appropriate conversion factor is chosen in order to convert from moles of the given substance
to moles of the unknown.
From the
balanced
equation
Known
Unknown
Known Mole &times; mole ratio
Mole
Mole
How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an
excess of nitrogen?
H2 = 4.20 mol
mol of NH3 = ?
4.20 mol H2 &times;
2 mol NH3
3 mol H2
= 2.80 mol NH3
12) One disadvantage of burning propane (C3H8) is that carbon dioxide (CO2) is one of the
products. The released carbon dioxide increases the concentration of CO 2 in the
atmosphere. How many moles of CO2 are produced when 10.0 mol of C3H8 are burned in
excess oxygen in a gas grill?
13) Methane and sulfur react to produce carbon disulfide (CS2), a liquid often used in the
production of cellophane.
CH4(g) +
S8(s) →
CS2(l) +
H2S(g)
a. Balance the equation.
b. Calculate the moles of CS2 produced when 1.50 mol S8 is used.
c. How many moles of H2S are produced?
pg. 7
Stoichiometry
Chemistry worksheets,
14) Sulfuric acid ( H2SO4 ) is formed when sulfur dioxide (SO2 ) reacts with oxygen and water.
a. Write the balanced chemical equation for the reaction.
b. How many moles of H2SO4 are produced from 12.5 moles of SO2?
c. How many moles of O2 are needed?
15) If you have 9 moles of H₂, how many moles of NH₃ can you make?
N₂+3H₂→2NH₃
2 H₂ + O₂ → 2 H₂O
16) If 8 moles of H2O are produced, how many moles of oxygen gas must be used?
17) If 14.00 moles of H₂ were used, how many moles of water would be produced?
18) What's the molar ratio between H₂SO₄ and H₂O
2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
19) If you have 30 moles of S₂ , how many FS₃ moles can you make?
2F + 3S₂ → 2FS₃
20) If you have 12 moles of NaClO₃ will produce how many moles of O₂?
2 NaClO3 → 2 NaCl + 3 O2
21)
To make 6 moles of AlCl3, how many moles of HCl would you need to start the reaction?
2Al+ 6HCl → 2AlCl₃ + 3H₂(g)
22) How many moles of aluminum sulfate will be produced from the reaction of 3 moles of
aluminum hydroxide?
2Al(OH)₂ + 3H₂SO₄ → Al₂(SO₄)₂ + 6H₂O
pg. 8
Stoichiometry
Chemistry worksheets,
Mole to mass conversion
In a chemical reaction if we know the number of moles for a component so we can calculate
the number of moles for all the other components and using the number of moles we can
calculate the mass for all the components.
From the
balanced
equation
Known
Unknown
Known Mole &times; mole ratio
Mole
From the
periodic
table
Mole
Moles &times;
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
𝟏𝒎𝒐𝒍𝒆
Mass
Determine the mass of sodium chloride (NaCl), commonly called table salt, produced when
1.25 mol of chlorine gas ( Cl2 ) reacts vigorously with excess sodium.
2Na(s) + Cl2(g) → 2NaCl(s)
moles of chlorine = 1.25 mol
Mole ratio:
2 mol NaCl
1 mol Cl 2
2 mol NaCl
Cl2
1.25 mol Cl2 &times;
mass of sodium chloride =? g
1mlo NaCl = 58.44 g
1 mol Cl 2
2.50 mol NaCl &times;
= 2.50 mol NaCl
58.44 g NaCl
1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
= 146 g NaCl
23) Sodium chloride is decomposed into the elements sodium and
chlorine by means of electrical energy. How much chlorine gas, in
grams, is obtained from the process diagrammed at right?
pg. 9
Stoichiometry
Chemistry worksheets,
24) Titanium is a transition metal used in many alloys because it is extremely strong and
lightweight. Titanium tetrachloride (TiCl4) is extracted from titanium oxide (TiO2) using
chlorine and coke (carbon).
TiO2 + C + 2Cl2 → TiCl4 + CO2
a. What mass of Cl2 gas is needed to react with 1.25 mol of TiO2?
b. What mass of C is needed to react with 1.25 mol of TiO2?
c. What is the mass of all of the products formed by reaction with 1.25 mol of TiO2?
25) What is the resulting Mg magnesium mass when 2.5mol of potassium K is reacted?
MgCl2(s) + 2 K(s) → Mg(s) + 2 KCl(s)
26) What is the hydrogen H2 mass resulting from the reaction of 4mol of zinc Zn?
Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
27) How many Aluminum Al Moles are produced from the 10.87 g Silver Ag reaction?
Al(NO3)3 + 3 Ag → Al + 3 AgNO3
pg. 10
Stoichiometry
Chemistry worksheets,
Mass to Moles Problems
Known
From the
periodic
table
Mass &times;
From the
balanced
equation
Unknown
Known Mole &times; mole ratio
Mole
Mole
𝟏𝒎𝒐𝒍𝒆
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
Mass
Acetylene gas, C2H2, is used in welding, produces an extremely hot flame when it burns in pure
oxygen according to the following reaction.
2 C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)
28)
How many moles of water (H2O) are produced when 25.0 grams of C2H2 burns completely?
3 Mg + 1 Fe2O3  2 Fe + 3 MgO
29) How many moles of iron, Fe, are produced with 25.0 grams of magnesium, Mg?
30) Laughing gas (nitrous oxide, N2O) is sometimes used as an anesthetic in dentistry.
NH4NO3(s)  N2O(g) + 2H2O(l)
a. How many moles of NH4NO3 are required to produce 33.0g of N2O?
b. How many moles of water are produced with 45.0g of N2O?
pg. 11
Stoichiometry
Chemistry worksheets,
Mass to Mass Problems
Known
From the
periodic
table
Mass &times;
From the
balanced
equation
Unknown
From the
periodic
table
Known Mole &times; mole ratio
Mole
Mole
𝟏𝒎𝒐𝒍𝒆
Moles &times;
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
𝟏𝒎𝒐𝒍𝒆
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
Mass
Mass
Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following
equation.
NH4NO3(s)→N2O(g)+2H2O(l)
31) In a certain experiment, 45.7 g of ammonium nitrate is decomposed. Find the mass of each
of the products formed.
given:
45.7 g NH4NO3
molar mass of NH4NO3 =
80.06 g/mol
molar mass of N2O = 44.02
g/mol
(mole ratios) 1 mol NH4NO3 = 1 mol N2O = 2 mol H2O
g NH4NO3 → mol NH4NO3 → mol 𝐍𝟐𝐎 → g 𝐍𝟐𝐎
4
Unknown:
mass N2O =? g
mass H2O =? g
3
0.6 mol N2O &times;
= 0.6 mol NH NO
4
80.06 g 𝑁𝐻4𝑁𝑂3
1 mol N2O
0.6mol NH4NO3 &times;
molar mass of H2O = 18.02
g/mol
1𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3
45.7 g NH NO &times;
1𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3
44.02 g N2O
1𝑚𝑜𝑙 N2O
= 0.6 mol N2O
1
= 25.1 g N2O
g NH4NO3 → mol NH4NO3 → mol H2O → g H2O
0.6mol NH4NO3 &times;
1.2 mol H O &times;
2
2 mol H2O
1𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3
18.02 g H2O
= 1.2 mol H2O
2
= 20.06 g H O
1𝑚𝑜𝑙 H2O
2
32) If 5.0 g of KClO₃ is decomposed, how many grams of KCl are produced?
2KClO₃ → 2KCl + 3O₂
pg. 12
3
(3.0 g)
Chemistry worksheets,
pg. 13
Stoichiometry
Stoichiometry
Chemistry worksheets,
33) How many potassium chloride grams are produced if 25 g of potassium chlorate
decompose? (15 g)
2KClO₃ → 2KCl + 3O₂
34) How many grams of hydrogen are necessary to react completely with 50.0 g of nitrogen in
the above reaction?
(10.8 g)
N₂ + 3H₂ → 2NH₃
35) How many grams of ammonia are produced when 50.0g of nitrogen reacts with hydrogen
in the above reaction?
(60.8 g)
36) How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with
an excess of barium chloride?
(4.2 g AgCl)
2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
37) How much BaCl₂ is necessary to react with 5.0 g of silver nitrate in the above reaction?
(3.1 g)
2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
pg. 14
Chemistry worksheets,
Stoichiometry
LiOH + HBr → LiBr + H₂O
38) If you start with 10.0 g of LiOH, how many grams of lithium bromide will be produced?
(36.25 g)
pg. 15
Stoichiometry
Chemistry worksheets,
Number of
particles
Number of
particles
𝟏𝒎𝒐𝒍𝒆
particles &times; 𝟔.𝟎𝟐𝟐&times;𝟏𝟎𝟐𝟑 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆
From the
periodic
table
Mass &times;
moles &times;
Known Mole &times; mole ratio
Mole
𝟏𝒎𝒐𝒍𝒆
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
Mass
From the
balanced
equation
𝟔.𝟎𝟐𝟐&times;𝟏𝟎𝟐𝟑 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆
𝟏𝒎𝒐𝒍𝒆
Mole
Moles &times;
From the
periodic
table
𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
𝟏𝒎𝒐𝒍𝒆
Mass
H₂SO₄ + 2NaOH → 2H₂O + Na₂SO₄
39) How many molecules of water are produced if 2.0 g of sodium sulfate are produced in the
above reaction? 2.41x10&sup2;⁴ molecules
2AlCl₃ → 2Al + 3Cl₂
40) If 10.0 g of aluminum chloride are decomposed, how many molecules of Cl₂ are produced?
6.77x10&sup2;&sup2; molecules
pg. 16
Stoichiometry
Chemistry worksheets,
Limiting Reactant
In the reaction,
N2(g)+3H2(g)→2NH3(g)
One mole of N2 will react with three moles of H2 to form two moles of NH3. Now let us suppose
that a chemist was to react three moles of N2 with six moles of H2.
six
of H2
three
moles
of N2
Hydrogen gas will be completely used up while there will be 1 mole of nitrogen gas left over
after the reaction is complete. Finally, the reaction will produce 4 moles of NH3 because that
is also two times as much as shown in the balanced equation. The overall reaction that
occurred in words:
2 mol N2 + 6 mol H2 → 4 mol NH3
All the amounts are doubled from the original balanced equation.
The limiting reactant (or limiting reagent) is the reactant that determines the amount of
product that can be formed in a chemical reaction.
The reaction proceeds until the limiting reactant is completely used up. In our example above,
the H2 is the limiting reactant.
The excess reactant (or excess reagent) is the reactant that is initially present in a greater
amount than will eventually be reacted.
In other words, there is always excess reactant left over after the reaction is complete. In the
above example, the N2 is the excess reactant.
pg. 17
Stoichiometry
Chemistry worksheets,
Determining the limiting reactant
sulfur dichloride is used to vulcanize rubber, a process that makes rubber harder, stronger, and
less likely to become soft when hot or brittle when cold. It is produced by reacting sulfur and
chlorine as follows:
S8(l) + 4Cl2(g) → 4S2Cl2(l)
41) Determine the mass (g) of S2Cl2 produced when 200.0g of S8 and 100.0g of Cl2 react.
Determine the limiting reactant, the excess reactant, the mass reacted and the excess
remaining.
100.0 g Cl2 &times;
given:
200.0g of S8
100.0g of Cl2
200.0 g S8 &times;
1𝑚𝑜𝑙 𝐶𝑙2
= 1.410 mol 𝐶𝑙2
70.91g 𝐶𝑙2
1𝑚𝑜𝑙 𝑆8
256.5 g 𝑆8
1.410 mol 𝐶𝑙2 available
=
= 0.7797 mol S8
1.808 mol 𝐶𝑙2 available
0.7797 mol 𝑆8 available
1 mol 𝑆8 available
From the balanced chemical equation each 1mol S8
need 4 moles Cl2 of but we have less amount of Cl2,
Unknown:
1. Determine the limiting
reactant
so Cl2is the limiting reactant
1.410 mol Cl2 &times;
2. Mass (g) of S2Cl2
S2Cl2
Analyze the excess reactant
1.410 mol Cl2 &times;
3. Mass reacted
0.3525 mol S8
4. Excess remaining
4 mol 𝐶𝑙2
1 mol 𝑆8
&times;
135.0 g 𝑆2𝐶𝑙2
1 mol 𝑆8𝐶𝑙2
= 190.4 g
2
= 0.3525 mol S8
4 mol 𝐶𝑙2
26505 g 𝑆8
3
= 90.42 g S8
1 mol 𝑆8
200.0 g S8 available - 90.42 g S8 needed = 109.6 g
S8 in excess
pg. 18
4 mol 𝑆2𝐶𝑙2
1
4
Chemistry worksheets,
Stoichiometry
42) The reaction between solid white phosphorus (P4) and oxygen produces solid tetraphosphorus
decoxide (P4O10). This compound is often called diphosphorus pentoxide because its empirical
formula is P2O5.
a. Determine the mass of P4O10 formed if 25.0 g of P4 and 50.0 g of oxygen are combined.
b. How much of the excess reactant remains after the reaction stops?
pg. 19
Chemistry worksheets,
Stoichiometry
43) The reaction between solid sodium and iron (III) oxide is one in a series of reactions that
inflates an automobile airbag: 6Na(s) + Fe2O3(s) → 3Na2O(s) + 2Fe. If 100.0 g of Na and
100.0 g of Fe2O3 are used in this reaction, determine the following.
a. limiting reactant
b. reactant in excess
c. mass of solid iron produced
d. mass of excess reactant that remains after the reaction is complete
pg. 20
Chemistry worksheets,
Stoichiometry
44) Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose
(C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for
photosynthesis.
a. Write the balanced chemical equation for the reaction.
b. Determine the limiting reactant.
c. Determine the excess reactant.
d. Determine the mass in excess.
e. Determine the mass of glucose produced
pg. 21
Chemistry worksheets,
Stoichiometry
Our bodies need vitamins, minerals, and some other elements in specific amounts to facilitate
normal metabolic reactions.
A lack of these substances can lead to
• Abnormalities in growth and development.
• Abnormalities in functioning of your body’s cells.
➢ Phosphorus is an essential element in living systems; phosphate groups occur regularly in
strands of DNA.
➢ Potassium is needed for proper nerve function, muscle control, and blood pressure. A diet
low
in potassium and high in sodium might be a factor in high blood pressure.
➢ Vitamin B-12. Without adequate vitamin B-12, the body is unable to synthesize DNA
properly, affecting the production of red blood cells.
Why do we use excess of a reactant?
We use one of the components of the reaction in abundant quantities and often the least
expensive reactant, this to ensure that the reaction is continuous until the limiting reactant
finished as well as to increase the speed of the reaction.
A Bunsen burner in the lab where the gas is burning with oxygen. The products depend on the
amount of oxygen available to react. If the oxygen is in excess amount, the flame color is blue,
and the carbon is completely burned. In case of lack of oxygen, the carbon does not burn
completely, which turns the color of flame to yellow and accumulates black soot on the surface
of the cup.
pg. 22
Stoichiometry
Chemistry worksheets,
Percent yield
Most reactions never succeed in producing the predicted amount of product. Reactions do not
go to completion or yield as expected for a variety of reasons.
1) Liquid reactants and products might adhere to the surfaces of their containers or
evaporate.
2) Products other than the intended ones might be formed by competing reactions, thus
reducing the yield of the desired product.
3) Some amount of any solid product is usually left behind on filter paper or lost in the
purification process.
We can easily calculate the amount of products in a particular reaction, but the important
question is, is the quantity of products really the same as the quantity we calculated? Chemical
reactions often do not produce the same mass values that we calculate in our calculations.
Hence, we have two values for the product mass, theoretical and actual yield.
Theoretical yield: The maximum amount of product that can be produced from specific
quantities of reactants.
Actual yield: The amount of product produced when a chemical reaction occurs during a
practical experiment.
Percent yield: the ratio of the actual yield to the theoretical yield.
percent yield =
pg. 23
(actual yield)
(theoretical yield)
&times; 100
Chemistry worksheets,
Stoichiometry
45) Solid silver chromate (Ag2CrO4) forms when excess potassium chromate (K2CrO4) is added
to a solution containing 0.500 g of silver nitrate (AgNO3). Determine the theoretical yield
of Ag2CrO4. Calculate the percent yield if the reaction yields 0.455 g of Ag2CrO4.
K2CrO4 + 2AgNO3 → Ag2CrO4 + 2KNO3
✓ Transfer the AgNO3 mass to the number of moles.
✓ Using the number of moles (AgNO3), calculate the number of moles (Ag2CrO4) produced.
✓ Using the number of moles (Ag2CrO4), calculate its mass (theoretical yield).
✓ Calculate the percent yield
46) Aluminum hydroxide (Al(OH)3) is often present in antacids to neutralize stomach acid (HCl).
The reaction occurs as follows: Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l). If 14.0 g of Al(OH)3 is
present in an antacid tablet, determine the theoretical yield of AlCl3 produced when the
tablet reacts with HCl.
pg. 24
Chemistry worksheets,
47) Zinc reacts with iodine in a synthesis reaction: Zn + I2 → ZnI2
a. Determine the theoretical yield if 1.912 mol of zinc is used.
b. Determine the percent yield if 515.6 g of product is recovered.
pg. 25
Stoichiometry
Chemistry worksheets,
Stoichiometry
48) When copper wire is placed into a silver nitrate solution (AgNO3), silver crystals and copper
(II) nitrate (Cu(NO3)2) solution form.
a. Write the balanced chemical equation for the reaction.
b. If a 20.0-g sample of copper is used, determine the theoretical yield of silver.
c. If 60.0 g of silver is recovered from the reaction, determine the percent yield of the
reaction.
pg. 26
Chemistry worksheets,
Stoichiometry
Percent Yield in the Marketplace
Sulfuric acid is an important chemical because it is a raw material used to make products such
as fertilizers, detergents, pigments, and textiles. The cost of sulfuric acid affects the cost of
many of the consumer items you use every day. The steps of the manufacturing process are
shown below.
Step 1
Step 2
Step 3
S8(s) + 8O2(g) → 8SO2(g)
2SO2(g) + O2(g) → 2SO3(g)
SO3(g) + H2O(l) → H2SO4(aq)
The first step, the combustion of sulfur, produces an almost 100% yield. The second step also
produces a high amount of products when the reaction takes place in the presence of a catalyst
and raises the temperature to 400 &deg; C
49) What is the catalyst?
50) What changes in the interaction with the temperature increase?
The reaction produces a thermal energy up to 600 &deg; C which reduces the amount of product.
In order for scientists to overcome this problem, the manufacturers re-pass the components
on a catalyst, cooling reaction and repeat the process four times, raising the percent yield up
to 98%.
pg. 27
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