Download Chapter 9 Stoichiometry

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Chemical reaction wikipedia , lookup

Transition state theory wikipedia , lookup

Electrochemistry wikipedia , lookup

Biochemistry wikipedia , lookup

Gaseous signaling molecules wikipedia , lookup

Rate equation wikipedia , lookup

Process chemistry wikipedia , lookup

Artificial photosynthesis wikipedia , lookup

Lewis acid catalysis wikipedia , lookup

Water splitting wikipedia , lookup

Isotopic labeling wikipedia , lookup

Redox wikipedia , lookup

Click chemistry wikipedia , lookup

Halogen wikipedia , lookup

Unbinilium wikipedia , lookup

Catalysis wikipedia , lookup

Gas chromatography–mass spectrometry wikipedia , lookup

PH wikipedia , lookup

Strychnine total synthesis wikipedia , lookup

Electrolysis of water wikipedia , lookup

Thermometric titration wikipedia , lookup

Vapor–liquid equilibrium wikipedia , lookup

Ununennium wikipedia , lookup

Atomic theory wikipedia , lookup

Bioorthogonal chemistry wikipedia , lookup

Stoichiometry wikipedia , lookup

Transcript
Stoichiometry
• Greek Translation
– Stoicheion – means “element”
– Metron – means “measure”
• Element Measuring
Stoichiometry
EQ: How do we compare elements so that
we can measure them?
For example, does…
23 grams of Na = 23 grams of Cl
23g Na x 1 mol = 1 mol Na or 6.022E23 atoms
23g
23g Cl x 1mol = 0.65 mol Cl or 3.85E23 atoms
35.5g
Stoichiometry
So how much Chlorine do I need since
NaCl is a one to one ratio?
23g Na x 1 mol = 1 mol Na or 6.022E23 atoms
23g
35.5g Cl x 1mol = 1 mol Cl or 6.022E23 atoms
35.5g
23g Na = 35.5g Cl
Stoichiometry
What if all I had was 23 grams of Chlorine?
How much NaCl could l make?
23g Na x 1 mol = 1 mol Na or 6.022E23 atoms
23g
23g Cl x 1mol = 0.65 mol Cl or 3.85E23 atoms
35.5g
Limiting Reagent
Chemical Stoichiometry
Stoichiometry - The study of quantities of
materials consumed and produced in
chemical reactions.
Consumed  Produced
Mole  Mole
Mass  Mole
Mole  Mass
Mass  Mass
Review: The Mole
 The number equal to the number of carbon atoms in
exactly 12 grams of pure 12C.
 1 mole of anything = 6.022 ´ 1023 units of that thing
Review: Molar Mass
A substance’s molar mass (molecular weight) is
the mass in grams of one mole of the compound.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
When the equation is balanced it has
quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen to
produce 2 moles of carbon dioxide and 3 moles of water
Mole to Mole Problems
Ammonia, NH3, is widely used as fertilizer and in many
household cleaners. How many moles of ammonia
are produced when 6 mol of hydrogen gas react
with an excess of nitrogen gas?
Step 1:
Step 2:
Step 3:
Step 4:
6 mol H2(g) X 2 mol NH3 =
3 mol H2
2=N=1
2=H=3
N2 + 3 H2  2NH3
4 mol NH3
mol unknown
mol known
Mole to Mole Problems
The decomposition of potassium
chlorate, KClO3, is used as a source of
oxygen in the laboratory. How many
moles of potassium chlorate are
needed to produce 15 mol of oxygen?
Mole to Mass Problems
In photosynthesis, plants use energy from the sun
to produce glucose, C6H12O6, and oxygen from
the reaction of carbon dioxide and water. What
mass, in grams, of glucose is produced when
3.00 mol of water react with carbon dioxide?
6CO2 + 6H2O  C6H12O6 + 6O2
3.00 mol H2O X 1 mol C6H12O6 X 180.18g =
6 mol H2O
1 mol
C6 X 12.01 = 72.06
H12 X 1.01 = 12.12
O6 X 16.00 = 96.00
180.18g/mol
90.09g C6H12O6
Mole-Mass Problem
Solution of iron (III) chloride, FeCl3, are used in
photoengraving and to make ink. This compound can
be made by the following reaction.
2 Fe + 3 Cl2 ---------> 2 FeCl3
How many grams of FeCl3 form from 2.4 moles of Fe?
Mass to Mole Problems
The first step in the industrial manufacture of nitric acid is the
catalytic oxidation of ammonia.
4 NH3 + 5 O2  4NO + 6H2O (unbalanced)
The reaction is run using 824 g of NH3 and excess oxygen.
a. How many moles of NO are formed?
b. How many moles of H2O are formed?
A. 824g NH3 X 1
mol X 4 mol NO = 48.39 mol NO
17.03g NH3
4 mol NH3
B. 824g NH3 X 1
mol X 6 mol H2O = 72.58 mol H2O
17.03g NH3
4 mol NH3
N X 14.00 = 14.00
H3 X 1.01 = 3.03
17.03g/mol
Mass to Mole Problems
Oxygen was discovered by Joseph Priestley in
1774 when he heated mercury (II) oxide to
decompose it to form its constituent elements.
a. How many moles of mercury (II) oxide are
needed to produce 125 g of oxygen?
b. How many moles of mercury are produced?
Mass to Mass Problems
When copper metal is added to silver nitrate in solution, silver
metal and copper (II) nitrate are produced. What mass of
silver is produced from 100. g of Cu?
Cu + 2 Ag(NO3)  2 Ag +
+2
-1
Cu(NO3)2
100g Cu X 1 mol Cu X 2 mol Ag X 107.87g = 339.48g Ag
63.55g
1 mol Cu
1 mol Ag
Molar Mass
Molar Mass
Mass to Mass Problems
Tin (II) fluoride is used in some toothpastes. It is
made by the reaction of tin (II) with hydrogen
fluoride according to the following equation.
Sn + 2 HF  SnF2 + H2
How many grams of SnF2 are produced from the
reaction of 30.00 g of HF with Sn?
Mole-Mole
How many moles of carbon dioxide, CO2, can be made
from Fe2O3 by the use of 1.90 moles of carbon
monoxide, CO, in the following reaction?
Fe2O3 + 3 CO -----------> 2 Fe + 3 CO2
Mole- Mass Problem
The Solvay process is used to make sodium carbonate, Na2CO3,
a chemical that ranked 11th among all chemicals in annual
production in 1986. The process begins with the passing of
ammonia and carbon dioxide through a solution of sodium
chloride. This makes sodium bicarbonate and ammonium
chloride: H2O + NaCl + NH3 + CO2 ---------> NH4Cl + NaHCO3
How many grams of sodium bicarbonate could, in theory, be
made from 5 moles of ammonia, NH3?
Mass-Mole
Using the equation: Pb(SO4)2 + 4 LiNO3  Pb(NO3)4 + 2 Li2SO4 ,
How many moles of lithium nitrate will be needed to make 250
grams of lithium sulfate, assuming that you have an adequate
amount of lead (IV) sulfate to do the reaction?
Mass to Mass Problems
What mass of aluminum is produced by
the decomposition of 5.0 kg of Al2O3?
Mole to Mole Problems
20 mol
X mol
CO2 + 2 LiOH  Li2CO3 + H2O
Mole Train
Use the mole ratio, Duh!
How many moles of lithium hydroxide are
required to react with 20 moles of carbon dioxide?
Given: 20 mol CO2
Unknown:
X mol LiOH
20 mol CO2
×
How can we go from moles of
one compound to mole of
another compound?
2 mol LiOH
1 mol CO2
=
40 mol LiOH
The MOLE TRAIN!!!!!!
How many moles of water are produced in the
complete combustion of 1.00 moles of glucose
according to the following reaction:
C6H12O6 + 6O2 ----> 6CO2 + 6H2O
6 mole H2O
1.00 moles C6H12O6 X -------------------1 moles C6H12O6
6.00 moles of water
What mass of H2SO4 can be formed from 5.0 moles of
SO2 and an unlimited quantity of H2O and O2?
2SO2 + O2 + 2H2O ---- 2H2SO4
2 moles H2SO4
5.0 moles SO2 X --------------- X
2 moles SO2
98 grams H2SO4
---------------------1 mole H2SO4
490 grams
The newly discovered element Takalahium (symbol Tak;
molecular mass = 411 g/mol) combines with oxygen to
form Takalahium Oxide. The unbalanced equation is:
Tak + O2 ---> Tak2O3 How many grams of Tak Oxide are
formed when burning 8.00 moles of Tak?
2 mole Tak2O3
8.00 moles Tak X --------------- X
4 mole Tak
870 grams Tak2O3
---------------------1 mole Tak2O3
3480 grams
A piece of zinc metal weighing 2.00 grams is
placed in an aqueous solution of silver nitrate. How
many moles of silver can be recovered from this
single replacement reaction?
• 1 Zn + 2 AgNO3  1 Zn(NO3)2 + 2 Ag
1 mole Zn
2 moles Ag
2.00 grams Zn X --------------- X
65.39 grams Zn
---------------------1 mole Zn
.061 moles
Mass- Mole Problem
The octane present in gasoline burns according to the
following equation:
2 C8H18 + 25 O2 ---------> 16 CO2 + 18 H2O
How many moles of CO2 are produced when 30 grams
of octane react with an excess of oxygen?
1 mole C8H18
30 grams C8H18 X --------------- X
114 grams C8H18
16 moles CO2
---------------------2 mole C8H18
2.10 moles
The formation of water from hydrogen and oxygen gas
is: 2H2 (g) + O2 (g) ---> 2H2O (l). What mass of water will
form from 12.0 grams of hydrogen and excess oxygen?
1 mole H2
2 mole H2O
18 grams H2O
12.0 grams H2 X --------------- X ------------------ X ---------------2 grams H2
2 moles H2
1 mole H2O
108 grams
In the reaction:
KClO3 + 5KCl + 6HNO3 --> 6KNO3 + 3Cl2 + 3H2O,
how many grams of KClO3 are required to
prepare 10.0 g of Cl2?
1 mole Cl2
1 mole KClO3
122.55 grams KClO3
10 grams Cl2 X --------------- X ------------------ X ----------------------70.9 grams Cl2 3 moles Cl2
1 mole KClO3
5.76 grams
Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al
6.50 g Al
+
3 O2  2Al2O3
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
= ? g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3
Limiting Reactant Video
The limiting reactant is the reactant that is
consumed first, limiting the amounts of
products formed.
Analogy
• You need to bake some cakes for the
school bake sale. The cake recipe calls
for 2 cups of flour and 4 eggs. If you
have 4 cups of flour and a dozen of
eggs, how many cakes can you make?
4 cups of flour X 2 cups flour
4 eggs
2 cups flour = 1 cake
4 eggs
= 2 cakes
Which ingredient ran out first (limiting)?
Silicon dioxide is usually quite unreactive but
reacts readily with hydrogen fluoride
according to the following equation.
4.5 mol SiO2
2.0 mol HF
SiO2 + 4 HF  SiF4 + 2 H2O
1 mol SiO2
4 mol HF
• If 2.0 mol of HF are exposed to 4.5 mol of
SiO2, which is the limiting reactant?
Look at the balanced equation and mole-to-mole ratio
How do you know which one to use?
If 12.0 g CH4 and 24.0 g O2 react to form CO2 and H2O,
how many grams of CO2 will be formed?
• 1 CH4 + 2 O2
1 mole CH4
 1 CO2 + 2 H2O
2 mole O2
32 grams O2
12 grams CH4 X --------------- X ------------------ X ----------------------16 grams CH4 1 mole CH4
1 mole O2
Need: 48 grams- O2 Limited: O2 will run out first
1 mole O2
1 mole CO2
44 grams CO2
24 grams O2 X --------------- X --------------- X ------------------32 grams O2 2 moles O2
1 mole CO2
16.5 grams CO2
Practice
Methane (CH4) reacts with O2 to form CO2 and water. If 30.0 grams of
CH4 and 64.0 grams of O2 are mixed and the reaction goes to
completion, how many grams of the reactant "in excess" remain?
• 1 CH4 + 2 O2
1 mole O2
 1 CO2 + 2 H2O
1 mole CH4
16 grams CH4
64 grams O2 X --------------- X ------------------ X -------------------32 grams O2
2 mole O2
1 mole CH4
Will Use 16 grams CH4
Excess Remaining = 30.0 grams – 16 grams =
14 grams
Some rocket engines use a mixture of hydrazine, N2H4, and
hydrogen peroxide, H2O2, as the propellant. The reaction is
given by the following equation.
N2H4 + 2 H2O2  N2 + 4 H2O
a. Which is the limiting reactant in this reaction when 0.750
mol of N2H4 is mixed with 0.500 mol of H2O2?
b. How much of the excess reactant, in moles, remains
unchanged?
c. How much of each product, in moles, is formed?
The black oxide of iron, Fe3O4, occurs in nature as the
mineral magnetite. This substance can also be made
in the laboratory by the reaction between red-hot iron
and steam according to the following equation.
3 Fe + 4 H2O  Fe3O4 + 4 H2
a. When 36.0 g of H2O is mixed with 167 g of Fe, which
is the limiting reactant?
b. What mass in grams of black iron oxide is produced?
c. What mass in grams of excess reactant remains when
the reaction is completed?
In the reaction PCl5 --> PCl3 + Cl2, 104.3 g of PCl5 are used,
and 30.0 g of Cl2 are produced. What is the % yield of Cl2 in
this reaction?
% Yield = Actual Yield
(given in problem)
X 100
Theoretical Yield (find with stoichiometry)
104.3 grams PCl5 X 1 mole
PCl5
208.25 grams PCl5
X 1 mole Cl2 X 70.9 grams Cl2
1 mole PCl5
1 mole Cl2
Theoretical Yield= 35.5 grams
Ans: 35.5
Actual Yield = 30.0 grams
Percent Yield =
30.0 grams Cl2 X 100
35.5 grams Cl2
Ans: 84.5 %
How many grams of CaF2 is formed
when 35.8mL of 0.678 M NaF is
treated with excess of Ca(NO3)2(aq)?
NaF + Ca(NO3)2  CaF2 + Na2(NO3)2
M = 0.678
35.8 mL X __1 L_ X 0.678 mol X 1 mol CaF2 X 78g CaF2
1000mL
1L
2 mol NaF
1 mol
0.947 g CaF2
Determine which compound is limiting
The reaction needs 14.4g of AgNO3 But we only have
6.00g. The reaction will stop when AgNO3 is gone
Now that you know which is the limiting reagent we can use it’s mass to
solve for how many grams of Ag2CO3
The End!
End!
The
End!
The
End!
The
End!