section 6.5—stoichiometry how can we determine in a lab the concentration of electrolytes?

Section 6.5—Stoichiometry

How can we determine in a lab the concentration of electrolytes?

2 H2 + O2 2 H2O

2

No coefficient = 1

2

For every 2 moles of H2…

1 mole of O2 is need to react…

and 2 moles of H2O are produced

What do those coefficients really mean?

What is stoichiometry?

Stoichiometry – Using the mole ratio from the balanced equation and information about one compound in the reaction to determine information about another compound in the equation.

Stoichiometry with Moles

Example:If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are

needed?2 H2 + O2 2 H2O

Stoichiometry with Moles

4.2 mole H2

mole H2

mole O2 = ________ mole O2

2

12.1

From balanced equation: 2 mole H2 1 mole O2

Example:If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are

needed?2 H2 + O2 2 H2O

But we can’t measure moles in lab!

We can’t go to the lab and count or measure moles…so we need a way to work in measurable units, such as grams and liters!

Molecular mass gives the grams = 1 mole of a compound!

Stoichiometry with Moles & Mass

Example:How many grams of AgCl will be

precipitated if 0.45 mole AgNO3 is reacted as follows:2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2

From balanced equation: 2 mole AgNO3 2 mole AgCl

Stoichiometry with Moles & Mass

0.45 mole AgNO3

mole AgNO3

mole AgCl = ________ g AgCl

2

265

Molar Mass of AgCl:1 mole AgCl = 143.35 g

mole AgCl

g AgCl

1

143.35

Example:How many grams of AgCl will be

precipitated if 0.45 mole AgNO3 is reacted as follows:2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2

Stoichiometry with Mass

Example:How many grams Ba(OH)2 are

precipitated from 14.5 g of NaOH in the following reaction:

2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl

From balanced equation: 2 mole NaOH 1 mole Ba(OH)2

Stoichiometry with Mass

14.5 g NaOH

g NaOH

mole NaOH

= ________ g Ba(OH)2

40.00

1

31.1

Molar Mass of NaOH:1 mole NaCl = 40.00 g

mole NaOH

mole Ba(OH)2

2

1

mole Ba(OH)2

g Ba(OH)2

1

171.35

Molar Mass of Ba(OH)2:1 mole Ba(OH)2 = 171.35 g

Example:How many grams Ba(OH)2 are

precipitated from 14.5 g of NaOH in the following reaction:


But what about for solutions?

Molarity gives the number of moles of the solute that are in 1 L of a solution

solutionL

solutemolesMolarity

Stoichiometry with Solutions

Example:If you need 15.7 g Ba(OH)2 to

precipitate, how many liters of 2.5 M NaOH solution is needed?



Stoichiometry with Solutions

15.7 g Ba(OH)2

g Ba(OH)2

mole Ba(OH)2

= ________ L NaOH

171.35

1

0.0733

Concentration of NaOH:2.5 mole NaOH = 1 L

mole Ba(OH)2

mole NaOH

1

2

mole NaOH

L NaOH

2.5

1


Example:If you need 15.7 g Ba(OH)2 to

precipitate, how many liters of 2.5 M NaOH solution is needed?


What about gases?

Standard Temperature and Pressure (STP) – 1 atm (or 101.3 kPa) and 273 K (0°C)

Molar Volume of a Gas – at STP, 1 mole of any gas = 22.4 liters

Stoichiometry with Gases

Example:If you need react 1.5 g of zinc completely,

what volume of gas will be produced at STP?

2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

From balanced equation: 1 mole Zn 1 mole H2

Stoichiometry with Gases

1.5 g Zn

g Zn

mole Zn

= ________ L H2

65.39

1

0.51

Molar volume of a gas:1 mole H2 = 22.4 L

mole Zn

mole H2

1

1

mole H2

L H2

1

22.4

Molar Mass of Zn:1 mole Zn = 65.39 g

Example:If you need react 1.5 g of zinc completely,

what volume of gas will be produced at STP?

2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

Keeping all these equalities straight!

TO GO BETWEEN USE THE EQUALITY

Grams & moles Molecular Mass in grams = 1 mole

moles & liters of a solution

Molarity in moles = 1 L

Moles & liters of a gas at STP

1 mole = 22.4 L at STP

2 different chemicals in a reaction

Coefficient ratio from balanced equation

Titrations—Using Stoichiometry

Titration – Addition of a known volume of a known concentration solution to a known volume of unknown concentration solution to determine the concentration.

End Point

End Point (or Stoichiometric Point) – When there is no reactant left over—they have all be reacted and the solution contains only products

Indicators – Paper or liquid that change color based on pH level.

The end point must be reached in order to use stoichiometry to calculate the unknown solution concentration

If the pH of the products is known, the indicator can be chosen to indicate the end point

Gravemetrics—Using Stoichiometry

Gravemetric Analysis – Using a reaction to precipitate out an insoluble compound. The solid is dried and massed. Stoichiometry can then be used to determine the original substance’s concentration from the mass of the precipitate

Let’s Practice #1

Example:If you are making 0.57 moles H2O, how many

moles of O2 are needed?2 H2 + O2 2 H2O

Let’s Practice #1

0.57 mole H2O

mole H2O

mole O2 = ________ mole O2

2

10.29

From balanced equation: 2 mole H2O 1 mole O2

Example:If you are making 0.57 moles H2O, how many

moles of O2 are needed?2 H2 + O2 2 H2O

Let’s Practice #2

Example:If you need to precipitate 10.7 g of

Ba(OH)2, how many grams NaOH are needed?



Let’s Practice #2

10.7 g Ba(OH)2

g Ba(OH)2

mole Ba(OH)2

= ________ g NaOH

171.35

1

5.00


mole Ba(OH)2

mole NaOH

1

2

mole NaOH

g NaOH

1

40.00

Molar Mass of NaOH:1 mole NaCl = 40.00 g

Example:If you need to precipitate 10.7 g of

Ba(OH)2, how many grams NaOH are needed?


Let’s Practice #3

Example:How many moles AgNO3 are needed to

react with 10.7 g CaCl2?2 AgNO3 + CaCl2 2 AgCl + 2 Ca(NO3)2

From balanced equation: 2 mole AgNO3 1 mole CaCl2

Let’s Practice #3

10.7 g CaCl2

g CaCl2

mole CaCl2 = ______ mole AgNO3

110.98

10.193

Molar Mass of CaCl2:1 mole CaCl2 = 110.98 g

mole CaCl2

mole AgNO3

1

2

Example:How many moles AgNO3 are needed to

react with 10.7 g CaCl2?2 AgNO3 + CaCl2 2 AgCl + 2 Ca(NO3)2

Let’s Practice #4

Example:How many liters of 0.10 M NaOH is

needed to react with 0.125 L of 0.25 M BaCl2?


From balanced equation: 2 mole NaOH 1 mole BaCl2

Let’s Practice #4

0.125 L BaCl2

L BaCl2

mole BaCl2

= ________ L NaOH

1

0.25

0.625

Concentration of NaOH:0.10 mole NaOH = 1 L

mole BaCl2

mole NaOH

1

2

mole NaOH

L NaOH

0.10

1

Concentration of BaCl2:0.25 mole BaCl2 = 1 L BaCl2

Example:How many liters of 0.10 M NaOH is

needed to react with 0.125 L of 0.25 M BaCl2?


Let’s Practice #5

Example:If you produce 15.4 L of H2 at STP, how many grams of ZnCl2 is also produced?2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

From balanced equation: 1 mole ZnCl2 1 mole H2

Let’s Practice #5

15.4 L H2

L H2

mole H2

= ________ g ZnCl2

22.4

1

93.7

Molar Mass of ZnCl2:1 mole ZnCl2 = 136.29 g

mole H2

mole ZnCl2

1

1

mole ZnCl2

g ZnCl2

1

136.29

Molar volume of a gas:1 mole H2 = 22.4 L

Example:If you produce 15.4 L of H2 at STP, how many grams of ZnCl2 is also produced?2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)

section 6.5—stoichiometry how can we determine in a lab the concentration of electrolytes?

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