STOICHIOMETRY
Chemical Quantities in Reactions
The total matter in the reactants of any chemical reaction is equal to the total
quantity of material in the goods. As a result, each reactant's total mass needs to be
equal to the combined mass of every product. The Law of Conservation of Mass, which
asserts that the total mass of the components interacting in a balanced chemical
reaction remains constant, is known as this. Therefore, when original substances are
transformed into new substances, no material is gained or lost. For instance, when
silver
combines
with
sulfur,
tarnish
(Ag2S)
is
created.
1.1 Mass Conservation
LEARNING GOAL: Determine the total mass of products and reactants in an
balance chemical formula.
The total matter in the reactants of any chemical reaction is equal to the total
quantity of material in the goods. As a result, each reactant's total mass needs to be
equal. to the combined mass of every product. The Law of Conservation of Mass is
the name given to this. It asserts that the overall mass of the substances reacting in
a balanced manner remains unchanged consequently when original chemicals are
transformed into novel materials. For instance, when silver combines with sulfur,
tarnish (Ag2S) is created.
2Ag(s) + S(s) → Ag2S(s)
2Ag(s)
S(s)
Ag2S(s)
Note: mass of the reactants 2Ag and S is the same as the mass of the product Ag2S
The number of silver atoms reacting in this reaction is twice that of sulfur atoms.
Atoms. 100 sulfur atoms are needed for every 200 silver atoms that react. Still, in
the real
chemical reaction, a significant number of silver and sulfur atoms would react. The
coefficients in the equation can then be understood in terms of moles when dealing
with molar amounts. Consequently, 1 mol of sulfur and 2 mol of silver combine to
form 1 mol of Ag2S. It is possible to calculate each one's mass and to express the
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moles of Ag, S, and Ag2S in terms of mass for each in grams. As a result, 247.9 g of
Ag2S is produced when 215.8 g of Ag and 32.1 g of S react.
Table 1: Information available from balanced equation:
Sample Problem: 1.1
Energy, water, and carbon dioxide are produced when oxygen and methane (CH4)
burn. When one mol of CH4 reacts, find the total mass of the reactants and products
for the following equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
The total mass of reactants and products is obtained by multiplying the
molar mass of each component by the coefficients in the equation, which represent
the number of moles of that substance. Because the coefficients in the balanced
equation are accurate, the mole quantities are also exact.
Solution:
SELF TEST:
Calculate the total mass of the reactants and the products for each of the following
equations:
a. 2Al(s) + 3Cl2(g) + 2AlCl3(s)
b. 4HCl(g) + O2(g) →2Cl2(g) + 2H2O(g)
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a. 2SO2(g) + O2(g) →2SO3(g)
b. 4P(s) + 5O2(g) →P2O5(s)
1.2 Mole Relationships in Chemical Equations:
LEARNING GOAL: Use a mole–mole factor from a balanced chemical equation to
Calculate the number of moles of another substance in the reaction.
Iron (III) sulfide is the result of an iron-sulfur reaction.
2 Fe(s) + 3S(s) → Fe2S3(s)
According to the balanced equation, two mol of iron and three mol of sulfur combine to form
Iron(III) sulfide in one mole. Actually, you can use any concentration of sulfur or iron, but the
ratio
Iron
2Fe(s)
Sulfur
+
3S(s)
Iron(III)Sulfide
→
Fe2S3(s)
In the chemical reaction of Fe and S, the mass of the reactants is the same as the mass of the product, Fe2S3.
of sulfur and iron combining will never change. The coefficients allow us to write
mole–mole ratios between reactants and products, as well as between reactants. The coefficients
The mole–mole factors employ precise numerical values, without restricting the quantity of
noteworthy numbers.
Using Mole–Mole Factors in Calculations
Every time you cook, tune an engine to run on the correct fuel and air mixture, or
In order to prepare medications in a pharmaceutical laboratory, you must be aware of how
much of the product will form and the appropriate amounts of reactants to utilize. Now that
we've completed writing every For the balanced equation 2Fe(s) + 3S(s)hFe2S3(s), we can
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find potential conversion factors. will utilize the mole–mole ratios in Sample Problem's
chemical computation.(1.2)
Sample Problem 1.2 : Calculating Moles of a Reactant:
In the chemical reaction of iron and sulfur, how many moles of sulfur are needed to
react with 1.42 moles of Iron?
2Fe(s) + 3S(s)→Fe2S3(s)
SOLUTION:
STEP I: State the given and needed quantities (moles).
STEP II: Write a plan to convert the given to the needed quantity (moles).
Moles of Fe x Mole – mole Factor x Mole of S
STEP III: Use coefficients to write mole–mole factors.
STEP IV: Set up the problem to give the needed quantity (moles).
SELF TEST:
For the following chemical equations
A. Write all of the mole–mole factors for each of the following chemical equations.
a. 2SO2(g) + O2(g)h2SO3(g)
b. 4P(s) + 5O2(g)h2P2O5(s)
B.
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Write the problem setup with the correct mole–mole factor needed to find each of the
following:
a. moles of SO3 from the moles of SO2
b. moles of O2 needed to react with moles of P
C. The chemical reaction of hydrogen with oxygen produces water.
2H2(g) + O2(g)→H2O(g)
a. How many moles of O2 are required to react with 2.6 mol of H2?
b. How many moles of H2 are needed to react with 5.0 mol of O2?
c. How many moles of H2O form when 2.5 mol of O2 reacts?
D. Carbon disulfide and carbon monoxide are produced when carbon is heated with
sulfur dioxide.
5C(s) + 2SO2(g) → CS2(l) + 4CO(g)
a. How many moles of C are needed to react with 0.500 mol of SO2?
b. How many moles of CO are produced when 1.2 mol of C reacts?
c. How many moles of SO2 are required to produce 0.50 mol of CS2?
E. In the acetylene torch, acetylene gas burns in oxygen to produce carbon dioxide,
water, and energy.
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
Acetylene
a. How many moles of O2 are needed to react with 2.40 mol of C2H2?
b. How many moles of CO2 are produced when 3.5 mol of C2H2 reacts?
c. How many moles of C2H2 are required to produce 0.50 mol of H2O?
d. How many moles of CO2 are produced when 0.100 mol of O2 reacts?
1.3: Mass Calculations for Chemical Reactions
LEARNING GOAL: Given the mass in grams of a substance in a chemical reaction,
Calculate the mass in grams of another substance in the reaction.
The mass of one can be used when the chemical reaction's balanced equation is available
of the reacting chemicals (A) to determine the mass of a different drug (B) in the response.
Nevertheless, the computations need that we translate A's mass into A's moles using
the mass of A in molars. The mole-mole factor that connects substance A to substance B is
then used, which comes from the balanced equation's coefficients. This factor of mole-mole
will convert A's moles into B's moles. Substance B's molar mass is utilized to Determine the
substance B's grams.
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Sample Problem 1.3: Calculating the Product:
When acetylene (C2H2) burns in oxygen, high temperatures are produced that are used
for welding metals. 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
How many grams of CO2 are produced when 54.6 g of C2H2 is burned?
SOLUTION:
STEP I. State the given and needed quantities (grams).
STEP II. Write a plan to convert the given to the needed quantity (grams).
STEP III. Use coefficients to write mole–mole factors; write molar masses.
STEP IV. Set up the problem to give the needed quantity (grams).
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SELF TEST:
A. When solid lead (II) sulfide reacts with oxygen gas, the products are solid lead(II) oxide
and sulfur dioxide gas.
a. Write the balanced chemical equation for the reaction.
b. How many grams of oxygen are required to react with 29.9 g of lead(II) sulfide?
c. How many grams of sulfur dioxide can be produced when 65.0 g of lead(II) sulfide reacts?
d. How many grams of lead (II) sulfide are used to produce 128 g of lead(II) oxide?
B. Nitrogen dioxide and water react to produce nitric acid, HNO3, and nitrogen oxide.
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
a. How many grams of H2O are required to react with 28.0 g of NO2?
b. How many grams of NO are produced from 15.8 g of H2O?
c. How many grams of HNO3 are produced from 8.25 g of NO2?
C. Calcium cyanamide, CaCN2, reacts with water to form calcium carbonate and ammonia.
CaCN2(s) + 3H2O(l) → CaCO3(s) + 2NH3(g)
a. How many grams of H2O are needed to react with 75.0 g of CaCN2?
b. How many grams of NH3 are produced from 5.24 g of CaCN2?
c. How many grams of CaCO3 form if 155 g of H2O reacts?
D. Iron(III) oxide reacts with carbon to give iron and carbon monoxide.
Fe2O3(s) + 3C(s)
→ 2Fe(s)
+ 3CO(g)
a. How many grams of C are required to react with 16.5 g of Fe2O3?
b. How many grams of CO are produced when 36.0 g of C reacts?
c. How many grams of Fe can be produced when 6.00 g of Fe2O3 reacts?
I.4: Calculating Mass of a Reactant
ss of a Reactant
The fuel heptane (C7H16) is designated as the zero point in the octane rating of gasoline.
Heptane is an undesirable compound in gasoline because it burns rapidly and causes
engine knocking. How many grams of O2 are required to react with 22.5 g of C7H16?
C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
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Solution:
STEP I. State the given and needed quantities (grams).
STEP II. Write a plan to convert the given to the needed quantity (grams).
STEP III. Use coefficients to write mole–mole factors; write molar masses.
STEP IV. Set up the problem to give the needed quantity (grams).
Self Test:
A.Nitrogen reacts with hydrogen to produce ammonia.
N2(g) + 3H2(g) → 2NH3(g)
a. If you have 3.64 g of H2, how many grams of NH3 can be produced?
b. How many grams of H2 are needed to react with 2.80 g of N2?
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c. How many grams of NH3 can be produced from 12.0 g of H2?
B. Ammonia and oxygen react to form nitrogen and water.
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)
a. How many grams of O2 are needed to react with 13.6 g of NH3?
b. How many grams of N2 can be produced when 6.50 g of O2 reacts?
c. How many grams of H2O are formed from the reaction of 34.0 g of NH3?
1.5: Limiting Reactants
LEARNING GOAL: Identify a limiting reactant when given the quantities of two
reactants; calculate the amount of product formed from the limiting reactant.
The reactants will typically not be present in the amounts suggested by the balancing
equation when a reaction is conducted. As a result, after the reaction is complete, some
reactants will be consumed and others will remain. The reactant that is used up first is
known as the limiting reactant. The maximum amount of product that can be created
depends on the concentration of the limiting reactant. The quantity of product generated,
known as the theoretical yield, is what would remain after the reaction has consumed all of
the limiting reactant. The highest amount of the product that may be produced from a specific
amount of limiting reactant is known as the theoretical yield. The actual yield, which is
typically less than the theoretical yield, is the amount of product that the reaction actually
produces. Chemists frequently compute the percent yield, which is defined as follows:
%yield = Actual yield/theoretical yield x 100%,
to convey the efficiency of a reaction.
Let's work through an example problem to better understand each of these ideas.
Sample Problem:
When we make peanut butter sandwiches for lunch, we need 2 slices of bread and 1
tablespoon of peanut butter for each sandwich. As an equation, we could write:
2 slices of bread + 1 tablespoon of peanut butter → peanut butter sandwich
We will run out of bread after eight slices if we have a full jar of peanut butter.
We have four sandwiches with peanut butter. We are unable to continue making sandwiches
after the bread is consumed, even if the jar still has a large amount of peanut butter. The quantity
of pieces of We can no longer prepare as many sandwiches with bread. We might have eight
slices of bread and only a spoonful of peanut butter on another day remaining. After we make
only a little, we will run out of peanut butter. I ate one peanut butter sandwich and had six pieces
of bread remaining. The tiny quantity of peanut butter supply has restricted the amount of
sandwiches we can make.
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The limiting reactant is the one that has been used up entirely. The excess reactant
is what remains after a complete reaction.
A. Calculating Moles of Product from a Limiting Reactant
Comparably, in a chemical reaction, the reactants don't necessarily mix in amounts that
Permit each to run out precisely at the same moment. A limiting reactant limits the
amount of product that can be generated in many reactions. Once we are aware of the
amounts. We determine the potential yield of a chemical reaction by calculating the
reactant-product ratio from each reactant in the event that it was fully consumed. The
reactant that runs out first and produces the least amount of product is the one we are
searching for.
Sample Problem:
Carbon monoxide and hydrogen are used to produce methanol (CH4O). The balanced
chemical reaction is
CO(g) + 2H2(g) → CH4O(g)
If 3.00 mol of CO and 5.00 mol of H2 are the initial reactants, what is the limiting reactant,
and how many moles of methanol can be produced?
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STEP I: State the given and needed quantity (moles).
STEP II: Write a plan to convert the quantity (moles) of each reactant to quantity (moles) of
product.
STEP III: Use coefficients to write mole–mole factors.
STEP IV: 4 Calculate the quantity (moles) of product from each reactant, and select the
smaller quantity (moles) as the limiting reactant.
Moles of CH4O (product) from CO:
The smaller amount, 2.50 mol of CH4O, is the maximum amount of methanol that can be
produced from the limiting reactant, H2, because it is completely consumed.
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Self Test:
A.A taxi company has 10 taxis.
a. On a certain day, only eight taxi drivers show up for work. How many taxis can be used
to pick up passengers?
b. On another day, 10 taxi drivers show up for work but three taxis are in the repair shop.
How many taxis can be driven?
B. Nitrogen and hydrogen react to form ammonia.
N2(g) + 3H2(g) → 2NH3(g)
Determine the limiting reactant in each of the following mixtures of reactants:
a. 3.0 mol of N2 and 5.0 mol of H2
b. 8.0 mol of N2 and 4.0 mol of H2
c. 3.0 mol of N2 and 12.0 mol of H2
B. Calculating Mass of Product from a Limiting Reactant
It is also possible to provide the reactant quantities in grams. The same computations must
be performed to determine the limiting reactant, but first the grams of each reactant must
be converted to moles, then to product moles, and lastly to grams of product. Next, choose
the smaller mass of the product that results from using the limiting reactant entirely.
Sample Problem: The Mass of the Product from a Limiting Reactant.
When silicon dioxide (sand) and carbon are heated, the products are silicon carbide, SiC,
and carbon monoxide. Silicon carbide is a ceramic material that tolerates extreme
temperatures and is used as an abrasive and in the brake discs of cars. How many grams of
CO are formed from 70.0 g of SiO2 and 50.0 g of C?
SiO2(s) + 3C(s) → SiC(s) + 2CO(g)
Solution:
STEP I. State the given and needed quantity in gram.
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STEP II. Write a plan to convert the quantity (grams) of each reactant to quantity (grams)
of product.
STEP III. Use coefficients to write mole–mole factors; write molar masses
STEP IV. Calculate the quantity (grams) of product from each reactant, and select the
smaller quantity (grams) as the limiting reactant.
Grams of CO (product) from SiO2:
The smaller amount, 65.3 g of CO, is the most CO that can be produced. This also means
that the SiO2 is the limiting reactant.
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Self Test:
A. For each of the following reactions, calculate the grams of indicated product when 25.0 g
of the first reactant and 40.0 g of the second reactant are used:
a. 2SO2(g) + O2(g) → 2SO3(g)
b. 3Fe(s) + 4H2O(l) → Fe3O4(s) + 4H2(g)
c. C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
B. For each of the following reactions, 20.0 g of each reactant is present initially.
Determine the limiting reactant, and calculate the grams of product in parentheses that
would be produced.
a. 2Al(s) + 3Cl2(g) →2AlCl3(s) + AlCl3
b. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
c. CS2(g) + 3O2(g) → CO2(g) + 2SO2(g)
C. For each of the following reactions, 20.0 g of each reactant is present initially. Determine
the limiting reactant, and calculate the grams of product in parentheses that would be
produced.
a. 4Al(s) + 3O2(g) → 2Al2O3(s) + (Al2O3)
b. 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
c. C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
1.5 PERCENT YIELD
LEARNING GOAL: Given the actual quantity of product, determine the percent yield for a
chemical reaction.
Up until now, we have assumed that every reactant in our issues has entirely transformed
to item. As a result, we determined the product's amount using the highest quantity that is
feasible. or 100 percent. Although it would be perfect, things don't generally work out this
way. While we transport as items are transferred between containers and a reaction is
carried out, some product is typically lost. The beginning ingredients used in the lab and in
commercial applications could not be entirely pure. and some of the reactants may be used
by side reactions to produce undesirable products. Consequently, 100% the intended
product is not truly acquired.
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In the lab, we measure very precise amounts when performing chemical reaction among
the reactants. We determine the reaction's theoretical yield, or the quantity if every
reactant was transformed into the intended product, we would anticipate a 100% yield of
product. The mass of the product, or the actual yield for the product, is collected and
measured once the reaction is complete. The real yield is lower than the theoretical yield
since some product is typically lost. We can get the percent yield for a product by combining
its actual and theoretical yields.
Percent yield (%) = actual yield/theoretical yield * 100%
Sample Problem: Calculating Percent Yield
On spacecraft, LiOH can be used to absorb exhaled CO2 from breathing air to form LiHCO3.
LiOH(s) + CO2(g) → LiHCO3(s)
What is the percent yield of LiHCO3 for the reaction if 50.0 g of LiOH gives 72.8 g of
LiHCO3?
Solution:
STEP I. State the given and needed quantities.
STEP II. Writer a plan to calculate the theoretical yield and the percent yield.
Percent yield (%) = actual yield of LiHCO3/ theoretical yield of LiHCO3* 100%
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STEP III. Use coefficients to write mole–mole factors; write molar masses.
STEP IV. Calculate the percent yield by dividing the actual yield (given) by the theoretical
yield and multiplying the result by 100%.
Calculation of percent yield:
A percent yield of 51.3% means that 72.8 g of the theoretical amount of 142 g of LiHCO3
was actually produced by the reaction.
Self Test :
A. When 30.0 g of carbon is heated with silicon dioxide, 28.2 g of carbon monoxide is
produced. What is the percent yield of carbon monoxide for this reaction?
B. When 56.6 g of calcium is reacted with nitrogen gas, 32.4 g of calcium nitride is
produced. What is the percent yield of Calcium nitride for this reaction?
3Ca(s) + N2(g) → Ca3N2(s)
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C. Carbon disulfide is produced by the reaction of carbon and sulfur dioxide.
5C(s) + 2SO2(g)→ CS2(g) + 4CO(g)
a. What is the percent yield of carbon disulfide if the reaction of 40.0 g of carbon
produces 36.0 g of carbon disulfide?
b. What is the percent yield of carbon disulfide if the reaction of 32.0 g of sulfur
dioxide produces 12.0 g of carbon disulfide?
Prepared by:
Mrs. Rosalia B. Cabalza
CAHS – MLS Faculty
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