STOICHIOMETRY

Interpreting Balanced Equations

Stoichiometrical Calculations

Limiting Reagents

Percent Yield

STOICHIOMETRY

Stoichiometry is the study of the amount

of substances produced and consumed in

chemical reactions.

The word stoichiometry comes from the

Greek “stoicheion” and “metrein” meaning

“element” and to “measure”.

Calculations are carried out step-by-step

using dimensional analysis.

INTERPRETING CHEMICAL

EQUATIONS

A balanced equation shows the ratios in

which the substances combine.

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)

atoms molecule Formula

units

molecule

mole moles mole moles

LSTP LSTP LSTP LSTP

g g g g

2 6 2 3

2 6 2 3

0 0 0

3(22.4)

= 67.2

2(26.98 g) =

53.96

6(36.46 g)

=218.76

2(133.33 g)

= 266.66

3(2.02 g)

=6.06

What is true about the mass of the products and the

mass of the reactants?

They are equal.

53.96 g + 218.76 g = 266.66 g + 6.06 g

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)

atoms molecule Formula

units

molecule

mole moles mole moles

LSTP LSTP LSTP LSTP

g g g g

2 6 2 3

2 6 2 3

0 0 0

3(22.4)

= 67.2

2(26.98 g) =

53.96

6(36.46 g)

=218.76

2(133.33 g)

= 266.66

3(2.02 g)

=6.06

According to the balanced equation, if two moles of aluminum react with 6

moles of hydrochloric acid, how many moles of aluminum chloride will be

produced? How many moles of hydrogen gas will be produced?

2 mol AlCl3 3 mol H2

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)

atoms molecule Formula

units

molecule

mole moles mole moles

LSTP LSTP LSTP LSTP

g g g g

2 6 2 3

2 6 2 3

0 0 0

3(22.4)

= 67.2

2(26.98 g) =

53.96

6(36.46 g)

=218.76

2(133.33 g)

= 266.66

3(2.02 g)

=6.06

How would the amount of products produced be different if

twice as many moles of each of the reactants were used?

The amount of products produced would also be doubled

(4 moles of AlCl3, 6 moles of H2)

2Al(s) + 6HCl(aq) → 2AlCl3(s) + 3H2(g)

A mole ratio is the ratio between the

numbers of moles of any two substances in a

balanced chemical equation.

Write the possible mole ratios for the reaction.

2 mol Al : 6 mol HCl

2 mol Al : 3 mol H2

2 mol Al : 2 mol AlCl3

6 mol HCl : 2 mol AlCl3

6 mol HCl : 3 mol H2 2 mol AlCl3 : 3 mol H2

Examples:

The mole ratio 2 mol Al: 6 mol HCl

Can be written as

Mole ratios can be written as conversion

factors.

2 mol Al

6 mol HCl or

6 mol HCl

2 mol Al

The mole ratio 6 mol HCl: 3 mol H2

Can be written as

6 mol HCl

3 mol H2

or 3 mol H2

6 mol HCl

DAILY STARTER #1

1. Write a sentence describing the following reaction in

terms of moles.

2NH3(g) N2(g) + 3H2(g)

2. Write the two conversion factors that could be

written for the relationship between moles of

ammonia and moles of hydrogen gas.

2 mol NH3

3 mol H2

or 3 mol H2

2 mol NH3

2 moles of ammonia gas decompose to form 1 mole of nitrogen gas

and 3 moles of hydrogen gas.

Stoichiometrical Calculations

A balanced chemical equation is necessary

in order to do stoichiometrical calculations

The Mole Bridge

Mass of

G = Mass of W

mol

of G

mol

of W

Volume

of G

(liters) at

STP

=

Volume of

W (liters)

at STP

aG bW

(given quantity) (wanted quantity)

G

G

massmolar

mol1

G

G

L 22.4

mol 1

Ga

Wb

mol

mol

W

W

mol 1

mass molar

W

W

mole 1

L 22.4

Example Problems. Remember to balance your

Equations!

1. How many moles of nitrogen gas are

required to completely react with 4.5

moles of hydrogen gas to produce

ammonia? N2 + 3H2 → 2NH3

2. Sodium chloride is prepared by the

reaction of sodium metal with chlorine

gas. How many moles of sodium are

required to produce 40.0 g of sodium

chloride?

2Na + Cl2 → 2NaCl

3. How many grams of aluminum will

be produced by the decomposition

of 25.0 g of aluminum oxide?

2Al2O3 → 4Al + 3O2

4. Oxygen is prepared in the laboratory by the

decomposition of potassium chlorate.

Calculate the volume of oxygen in liters,

measured at STP, that would be obtained

from 183.9 g of potassium chlorate.

2KClO3 → 2KCl + 3O2

5. Calculate the volume of carbon dioxide

that will be produced at STP from the

complete combustion of 6.0 L of ethane

gas (C2H6).

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

6. If 30.0 g of lithium react with an

excess of water, how many grams of

lithium hydroxide will be produced?

2Li + 2H2O → 2LiOH + H2

DAILY STARTER #2

2NH3(g) N2(g) + 3H2(g)

1. What volume of hydrogen gas will be produced at

STP by the decomposition of 45.0 g of ammonia?

2. What mass of ammonia is needed to produce

3.50×1023 molecules of nitrogen gas?

45.0 g NH3 ×1 mol NH3

17.04 g NH3

×3 mol H2

2 mol NH3

×22.4 L H2

1 mol H2

= 88.7 L H2

3.50 × 1023molecules N2 ×1 mol N2

6.02 × 1023 molecules N2

×2 mol NH3

1 mol N2

×17.04 g NH2

1 mol NH3

= 19.8 L NH3

LIMITING REAGENT

The limiting reagent in a chemical

reaction limits the amounts of the

other reactants that can combine –

and the amount of product that can

form – in a chemical reaction.

The reaction will stop once the

limiting reagent runs out.

EXCESS REAGENT

The excess reagent in a chemical

reaction is the substance that is not

used up completely in a chemical

reaction.

AN EXAMPLE

Mrs. Kyle likes to make grilled cheese sandwiches for her

children.

2 slices bread + 1 slice cheese + 1 T Butter + energy → 1 sandwich

How many sandwiches could Mrs. Kyle make if she has 10

slices of bread, 4 slices of cheese and 7 tablespoons of

butter? Why?

Mrs. Kyle could only make 4 sandwiches because she only

has enough cheese for 4 sandwiches.

What left over ingredients would Mrs. Kyle have?

Mrs. Kyle would have 2 slices of bread and 3 tablespoons of

butter left over.

Ammonia is prepared by the reaction of hydrogen

gas with nitrogen gas.

3H2 + N

2 → 2NH

3

Determine the maximum amount of ammonia (in moles)

that could be produced when 6.0 mol of H2 reacts with

4.0 mol of N2 and identify the limiting reagent.

The maximum amount of NH3 that could be

produced is 4.0 mol and H2 is limiting.

6.0 mol H2 ×2 mol NH3

3 mol H2

= 4.0 mol NH3

4.0 mol N2 ×2 mol NH3

1 mol N2= 8.0 mol NH3

Ammonia is prepared by the reaction of hydrogen gas

with nitrogen gas.

3H2 + N

2 → 2NH

3

How many moles of the excess reagent remain

unreacted?

4.0 mol N2 – 2.0 mol N2 = 2.00 mol N2

unreacted

6.0 mol H2 ×1 mol N2

3 mol H2

= 2.0 mol N2reacted

Ammonia is prepared by the reaction of hydrogen gas

with nitrogen gas.

3H2 + N

2 → 2NH

3

Draw a representation of the reaction before and after

the reaction.

Before Reaction After Reaction

H2

N2

NH3

Methanol, CH3OH, is the simplest of the alcohols. It is

synthesized by the reaction of hydrogen and carbon

monoxide.

2H2

+ CO → CH3OH

If 150 mol of carbon monoxide and 400 moles of

hydrogen chemically combine, determine the maximum

amount (in mol) of methanol that could be produced and

identify the limiting reagent.

The maximum amount of CH3OH that could be

produced is 150 mol and CO is limiting.

Methanol, CH3OH, is the simplest of the alcohols. It is

synthesized by the reaction of hydrogen and carbon

monoxide.

2H2

+ CO → CH3OH

How many mol of the excess reagent remain

unreacted?

400 mol H2 – 300 mol H2 = 100 mol H2

A student reacts 80.0 g of copper metal with 25.0 g of sulfur

to form copper(i) sulfide.

2Cu + S → Cu

2S

Determine the maximum amount of copper(I) sulfide (in

grams) that could be produced and identify the limiting

reagent.

The maximum amount of Cu2S that can be produced is 100. g. The

limiting reagent is copper.

A student reacts 80.0 g of copper metal with 25.0 g of sulfur

to form copper(i) sulfide.

2Cu + S → Cu

2S

How many grams of the excess reagent remain unreacted?

25.00 g S – 20.19 g S = 4.81g S

DAILY STARTER #3

Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

1. Identify the limiting reagent when 20.0 g of Mg react

with 20.0 g of HCl.

45.0 g NH3 ×1 mol NH3

17.04 g NH3

×3 mol H2

2 mol NH3

×22.4 L H2

1 mol H2

= 88.7 L H2

3.50 × 1023molecules N2 ×1 mol N2

6.02 × 1023 molecules N2

×2 mol NH3

1 mol N2

×17.04 g NH2

1 mol NH3

= 19.8 L NH3

DAILY STARTER #3

2. Aluminum metal reacts with oxygen gas to produce

aluminum oxide.

a. Determine the maximum mass of aluminum oxide that

could be produced when 20.0 g of aluminum reacts

with 30.0 g of oxygen gas and identify the limiting

reactant.

20.0 g Al ×1 mol Al

26.98 g Al×

2 mol Al2O3

4 mol Al×

101.96 g Al2O3

1 mol Al2O3= 37.8 L Al2O3

30.0 g O2 ×1 mol O2

32.00 g O2

×2 mol Al2O3

3 mol O2

×101.96 g Al2O3

1 mol Al2O3= 63.7 Al2O3

4Al(s) + 3O2(g) 2Al2O3(s)

The maximum mass of aluminum oxide that could be produced is 37.8 g and the

limiting reactant is aluminum.

DAILY STARTER #3

2. Aluminum metal reacts with oxygen gas to produce

aluminum oxide.

b. Determine the mass of excess reagent remaining.

20.0 g Al ×1 mol Al

26.98 g Al×

3 mol O2

4 mol Al×

32.00 g O2

1 mol O2= 17.8 g O2 reacted

4Al(s) + 3O2(g) 2Al2O3(s)

30.0 g O2 – 17.8 g O2 = 12.2 g O2

PERCENT YIELD

The theoretical yield is the calculated

amount of product that could form

during a reaction based upon a

balanced equation.

This is the maximum amount of

product that could be formed.

PERCENT YIELD

The actual yield is the amount of

product that forms when the reaction is

carried out in the laboratory.

The actual yield is often less than the

theoretical yield.

PERCENT YIELD

The percent yield is the ratio of the

actual yield to the theoretical yield.

WHAT ARE SOME FACTORS THAT MIGHT

CAUSE THE PERCENT YIELD TO BE LESS

THAN 100%?

1. Reactions do not always go to

completion.

2. Impure reactants and competing side

reactions may cause other products to

be formed.

3. Some of the product may be lost

during purification.

WHAT ARE SOME FACTORS THAT MIGHT

CAUSE THE PERCENT YIELD TO APPEAR

TO BE MORE THAN 100%?

1. The product might not be completely

dry.

2. The product might combine with

oxygen in the air drying it.

A student decomposed 24.8 g calcium carbonate in

the laboratory.

CaCO3 → CaO + CO

2

Calculate the theoretical yield of calcium oxide.

What is the percent yield if the student only produced

13.1 g of calcium oxide in the laboratory?

When 50.0 g of silicon dioxide is heated with an excess

of carbon, 32.2 g of silicon carbide is produced.

SiO2(s) + 3C(s) → SiC(s) + 2CO(g)

Calculate the theoretical yield.

Calculate the percent yield.

Aluminum reacts with excess copper(II) sulfate according to

the reaction given below. If 1.95 g of Al react and the percent

yield of Cu is 60.0%, what mass of cu is produced?

2Al(s) + 3CuSO4(aq)

→ Al

2(SO

4)3(aq) + 3Cu(s)

First calculate the theoretical yield of copper.

Next calculate the mass of Cu produced.

Aluminum reacts with excess copper(II) sulfate according to

the reaction given below. If 1.95 g of Al react and the percent

yield of Cu is 60.0%, what mass of Cu is produced?

2Al(s) + 3CuSO4(aq)

→ Al

2(SO

4)3(aq) + 3Cu(s)

First calculate the theoretical yield of copper.

Next calculate the mass of Cu produced.