Upload
winfred-powell
View
215
Download
0
Embed Size (px)
Citation preview
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:
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
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?
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
From balanced equation: 2 mole NaOH 1 mole Ba(OH)2
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
Molar Mass of Ba(OH)2:1 mole Ba(OH)2 = 171.35 g
Example:If you need 15.7 g Ba(OH)2 to
precipitate, how many liters of 2.5 M NaOH solution is needed?
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
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?
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
From balanced equation: 2 mole NaOH 1 mole Ba(OH)2
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
Molar Mass of Ba(OH)2:1 mole Ba(OH)2 = 171.35 g
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?
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
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?
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
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?
2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl
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)