Stoichiometry

StoichiometryStoichiometry

Consider the chemical equation:4NH4NH33 + 5O + 5O22 6H 6H22O + 4NOO + 4NO

There are several numbers involved. What do they all mean?

““stochio” = elementstochio” = element““metry” = measurementmetry” = measurement

  Stoichiometry is about measuring the amounts of Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction. elements and compounds involved in a reaction.

StoichiometryStoichiometry

4NH4NH33 + 5O + 5O22 6H 6H22O + 4NOO + 4NO

4 : 5 : 6 : 4is the mole ratio.

StoichiometryStoichiometry

4NH4NH33 + 5O + 5O22 6H 6H22O + 4NOO + 4NO

4 moles of NH3 react with 5 moles of O2

to produce 6 moles of H2O and 4 moles of NO

5

Stoichiometry (more working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)2 2H2O + BaCl2 1 1

2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

coefficients give MOLAR RATIOS

STOICHIOMETRY

MASS A VOLUME A

MOL “B”

MASS B VOLUME B

MOL “A”

Moving along the stoichiometry path

We always use the same type of information to make the jumps between steps:

grams (x) moles (x) moles (y) grams (y)

Molar mass of x Molar mass of y

Mole ratio from balanced equation

© D Scott; CHS

Many stoichiometry problems follow a pattern:

grams(x) moles(x) moles(y) grams(y)

Converting grams to grams

We can start anywhere along this path depending on the question we want to answer

Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles.

© D Scott; CHS

Calculating Masses of Calculating Masses of Reactants and ProductsReactants and Products

1.1. Balance the equation.Balance the equation.

2.2. Convert mass or volume to Convert mass or volume to moles, if necessary.moles, if necessary.

3.3. Set up mole ratios.Set up mole ratios.

4.4. Use mole ratios to calculate Use mole ratios to calculate moles of desired material.moles of desired material.

5.5. Convert moles to mass or Convert moles to mass or volume, if necessary.volume, if necessary.

1.1. Balance the equation.Balance the equation.

2.2. Convert mass or volume to Convert mass or volume to moles, if necessary.moles, if necessary.

3.3. Set up mole ratios.Set up mole ratios.

4.4. Use mole ratios to calculate Use mole ratios to calculate moles of desired material.moles of desired material.

5.5. Convert moles to mass or Convert moles to mass or volume, if necessary.volume, if necessary.

Mole Ratios

A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

11

When N2O5 is heated, it decomposes:

2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5

52

2

ON mol2

NO mol48.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O2

4.3 mol N2O5

52

2

ON 2mol

O mol12.2

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

2N2O5(g) 4NO2(g) + O2(g)4.3 mol ? mol

Mole – Mole Conversions

Units match

12

When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5

210 g NO2

2

52

NO mol4

ON mol22.28

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O5

75.0 g O2

2

52

O 1mol

ON mol2506

2

2

NO g0.46

NO mol

2

2

O g 32.0

O mol

52

52

ON mol

ON g108

gram ↔ mole and gram ↔ gram conversions

2N2O5(g) 4NO2(g) + O2(g)210g? moles

2N2O5(g) 4NO2(g) + O2(g)75.0 g? grams

Units match

Example

Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks)

2 Mg(s) + O2(g) 2 MgO(s)

Mole Ratios:

2 : 1 : 2

How many moles of MgO is produced if 4 mols of Mg reacts with unlimited supply of O2

1) N2 + 3 H2 ---> 2 NH3

a) How many moles of NH3 if 6 moles of H2 reacts with unlimited supply of N2?

b) How many moles of NH3 is produced if 5 moles of H2 reacts with unlimited supply of N2?

Practise Problems

4NH3 + 5O2 6H2O + 4NO

How many moles of H2O are produced if 2.00 moles of O2 are used?

Stoichiometry Question (1)

2.00 mol O2 2.40 mol H2O=

Notice that a correctly balanced equation is essential to get the right answer

6 mol H2O

5 mol O2

© D Scott; CHS

4 mol NO

6 mol H2O

4 NH3 + 5 O2 6 H2O + 4 NO

How many moles of NO are produced in the reaction if 15 mol of H2O are also produced?

Stoichiometry Question (2)

15 mol H2O 10. mol NO=

© D Scott; CHS

18.02 g H2O

1 mol H2O

6 mol H2O

4 mol NH3

4 NH3 + 5 O2 6 H2O + 4 NO

How many grams of H2O are produced if 2.2 mol of NH3 are combined with excess oxygen?

Stoichiometry Question (3)

2.2 mol NH3 59 g H2O

=

© D Scott; CHS

5 mol O2

6 mol H2O

32 g O2

1 mol O2

4 NH3 + 5 O2 6 H2O + 4 NO

How many grams of O2 are required to produce 0.3 mol of H2O?

Stoichiometry Question (4)

0.3 mol H2O 8 g O2=

© D Scott; CHS

4 NH3 + 5 O2 6 H2O + 4 NO

How many grams of NO is produced if 12 g of O2 is combined with excess ammonia?

4 mol NO

5 mol O2

x

Stoichiometry Question (5)

12 g O2

9.0 g NO=

30.01 g NO

1 mol NOx

1 mol O2

32 g O2

x

© D Scott; CHS

Have we learned it yet?Try these on your own - 4 NH3 + 5 O2 6 H2O + 4 NO

a) How many moles of H2O can be made using 1.6 mol NH3?

b) what mass of NH3 is needed to make 0.75 mol NO?

c) how many grams of NO can be made from 47 g of NH3?

© D Scott; CHS

4 NH3 + 5 O2 6 H2O + 4 NO

a)

b)

c)

Answers

6 mol H2O

4 mol NH3

x 1.6 mol NH3 2.4 mol H2O

=

4 mol NH3

4 mol NOx 0.75 mol NO 13 g

NH3

= 17.04 g NH3

1 mol NH3

x

4 mol NO

4 mol NH3

x 47 g NH3

83 g NO=

30.01 g NO

1 mol NOx

1 mol NH3

17.04 g NH3

x

© D Scott; CHS

Mass-Mass Stoichiometry

23

How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

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

Gram to Gram Conversions

24

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

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

Now let’s get organized. Write the information below the substances.

3.45 g ? grams

Gram to Gram Conversions

25

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams

Let’s work the problem.

= g AlCl3

3.45 g Al

Alg 27.0

Almol

We must always convert to moles.

Now use the molar ratio.

Almol 2

AlClmol 2 3

Now use the molar mass to convert to grams.

3

3

AlClmol

AlClg 133.317.0

Units match

gram to gram conversions

Mass-Volume/Concentration Stoichiometry

Just follow mass-mass problem to the penultimate level

Convert moles of the substance into Volume.

Volume-Volume Stoichiometry

Just follow mass-mass problem to the penultimate level

Convert moles of the substance into Volume.

28

50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?

H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

29

50.0 mL

6.0 M

L

mol 6.0

? g

50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?

H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

=

Our Goal

= g NaHCO3

H2SO4

50.0 mL

1000mL

SOH mol 6.0

42SOH

42

1 molH2SO4

NaHCO3

2 molNaHCO3

84.0 gmolNaHCO3

50.4

30

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

First write a balancedEquation.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

31

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now, let’s get organized. Place numerical Information and

accompanying UNITS below each compound.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 ML

mol

? mL

35.0 mL

mL 1000

mol 0.125

L

mol 0.125

Since 1 L = 1000 mL, we can use this to save on the number of conversions

Our Goal

32

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now let’s get to work converting.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 ML

mol

? mL

35.0 mL

mL1000

mol 0.125

L

mol 0.125

= mL NaOH

H2SO4

35.0 mL H2SO4

0.125 mol 1000 mL H2SO4

NaOH2 mol1 mol H2SO4

1000 mL NaOH0.102 mol NaOH

85.8

Units Match

Solution Stoichiometry:

shortcut

33

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

1st write out a balanced chemical

equation

Solution Stoichiometry

34

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

2HCl(aq) + Ba(OH)2(aq) 2H2O(l) + BaCl2

0.40 M 47.1 mL0.75 M? mL

= mL HCl

Ba(OH)2

47.1 mL

2

2

Ba(OH)

Ba(OH)

mL 1000

0.75mol

1 mol Ba(OH)2

HCl2 mol

0.40 mol HCl

HCl1000 mL

176

Units match

Solution Stoichiometry

35

Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

First write a balanced chemical reaction.

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1

23.28 mL

0.135 mol L

25.00 mL

? mol L

36

Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1

23.28 mL

0.135 mol L

25.00 mL

? mol L

= mol Ba(OH)2

L Ba(OH)2

25.00 x 10-3 L Ba(OH)2

Units Already Match on Bottom!

HClmL 23.28

HCl

HCl

mL 1000

mol 0.135

HCl

Ba(OH)

mol 2

mol l2 0.0629

Units match on top!

37

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

We must first write a balanced equation.

Solution Stochiometry Problem:

38

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

Ca(OH)2(aq) + HNO3(aq) H2O(l) + Ca(NO3)2(aq)2 248.0 mL 19.2 mL

0.385 ML

mol 0.385

= mol(Ca(OH)2)

L (Ca(OH)2)

19.2 mLHNO3

3

3

HNO

HNO

mL 1000

mol0.385

3

2

HNO 2mol

Ca(OH) 1mol

48.0 x 10-3L

? M

units match!

0.0770

Solution Stochiometry Problem: