Stoichiometry Chapter 11 (page 326). Essential Question!! Why do scientist compare mole ratios, and...
Transcript of Stoichiometry Chapter 11 (page 326). Essential Question!! Why do scientist compare mole ratios, and...
Stoichiometry
Chapter 11(page 326)
Essential Question!!
• Why do scientist compare mole ratios, and why is it important in chemical reactions??
Vocabulary: Section 1
• Stoichiometry• Moles• Stoichiometric equivalent• Mole ratio• Gram to gram calculations
Chemical equations tell stories…
But what exactly do they tell us?
2CO(g) + O2(g) 2CO2(g)
They tell us what compounds we start with:Carbon monoxide (CO) gas
Oxygen (O2) gas
what compounds are formed:Carbon dioxide (CO2) gas
Chemical equations tell stories…
What else do they tell us?
2CO(g) + O2(g) 2CO2(g)
stoichiometry: the study of the amounts of substances involved in a chemical reaction.
2 CO molecules 2 CO2 molecules1 O2 molecules
They tell us how much of each compound is involved
2CO(g) + O2(g) 2CO2(g)
2 CO molecules2 dozen CO molecules2 moles CO molecules
2 x (6.023 x 1023) CO molecules
1 O2 molecules1 dozen O2 molecules
1 mole O2 molecules(1 x) 6.023 x 1023 O2 molecules
2 CO2 molecules2 dozen CO2 molecules2 moles CO2 molecules
2 x (6.023 x 1023) CO2 molecules
2CO(g) + O2(g) 2CO2(g)
2 moles
CO molecules
1 mole
O2 molecules
2 moles
CO2 molecules
Number of moles is not conserved
Is that okay?
≠+
2CO(g) + O2(g) 2CO2(g)
2 C atoms
2 O atoms2 O atoms
2 C atoms
4 O atoms
Number of atoms is
conserved
This chemical equation is balanced
=
=+
Write as a ratio:
Coefficients are important
+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes1 bagcake mix
1 4
4 1
cup oil batches cupcakesor
batches cupcakes cup oil
Coefficients are important
C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)
Fermentation of sugar (glucose) into alcohol:
1 moleglucose
2 molesethanol
2 molescarbon dioxide
1 moles
3 moles
7.5 moles
glucose will yield
2 moles
6 moles
15 moles
ethanol/ carbon dioxide
x 7.5x 3
C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)
Fermentation of sugar (glucose) into alcohol:
1 moleglucose
2 molesethanol
2 molescarbon dioxide
Write as a ratio: 1 2
2 1
mole glucose moles ethanol
moles ethanol mole glucose
mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.
CO(g) + 2H2(g) CH3OH(l)
Mole ratios
Consider the following equation:
carbon monoxide
hydrogen methanol
3
3
11
1 1
molemole
m
CH
ol
OHCO
e mCH OH Oole C
Compare the reactant CO to the product CH3OH.
Stoichiometry1. Mass of substance A
2. Amount in mol of
Substance A
3. Amount in mol of
substance B
4. Mass of Substance B
Convert by dividing by the molar mass of A
Convert by mult. by molar mass of B
Convert using
the mol ratio
given in the chemical equation
B
A
Process for calculating grams from grams given
Introduction to Stoichiometry
1. mole --> molemoles of A X coeff B
coeff A= moles B
2. Mole--> massmoles of A X coeff B
coeff AX molar mass
of B = mass of B
Solution Route
3. mass --> molemass A FM A X
coeff Bcoeff A
= mol B
4. mass --> massmass AFM A X coeff B
coeff A X FM B = mass B
Method Mass - MassMethod Mass - Mass
mass Amass Acoeff A X FM A coeff A X FM A ==
mass Bmass B
coeff B X FM Bcoeff B X FM B
Method Mass - VolumeMethod Mass - Volume
mass Amass Acoeff A * FM A coeff A * FM A
== vol Bvol Bcoeff B * 22.4coeff B * 22.4
Method Volume - VolumeMethod Volume - Volume
Vol AVol Acoeff Acoeff A
** vol Bvol Bcoeff Bcoeff B
Assignment
• Take a new sheet of paper and fold it into three sections
• Write your name, the title of the chapter and the number
• On the first section from the sheet of paper, please write six things that you learned from your notes so far that could appear on your test.
1.Given the equation N2 + 3 H2 2NH3 If 4.00 mol of H2 react, how many mol of NH3 will be produced?
4.0 mol X23
= 2.66 mol
How many moles of sodium will react with water to produce 4.00 mol of Hydrogen?
2 Na + 2H2O 2 NaOH + H2
4 mol H2 X2
18 mol sodium=
How many moles of H2SO4 will react with 18.0 mol of Al?
2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2
18.0 mol Al X 32
= 27.0 moles
What mass of KClO3 do you need to produce 0.50 mol of O2?
2 KClO3 2 KCl + 3 O2
0.50 mol X 23
= 0.33 mol KClO3
0.33 mol KClO3 X 122.6 g/mol
= 40.866 g
What mass of Zn metal do you need to produce 0.50 mol of ZnCl2?
Zn (s) + 2HCl ZnCl2(s) + H2(g)
0.50 mol X 11
= 0.5 Zn
0.5 mol Zn X 65.4 g/mol
= 32.7 g Zn
What mass of NH3 do you need to produce 0.25 mol of H2?
N2 + 3H2(g) 2 NH3
0.25 mol X 32
= 0.375 H2
0.375 mol H2 X 0.75 g/mol
= 0.75 g H2
Assignment
• Write a three dollar summary of this section (based on what you learned) and using three vocabulary words learned
• Answer questions # 1- 3 on page 360• Turn in completed work
• Honors Chemistry Homework:– Page 362 # 38 - 45
Vocabulary: Section 2
• Percent yield• Actual yield• Theoretical yield
In theory, all 100 kernels should have popped.
Did you do something wrong?
+
100 kernels 82 popped 18 unpopped
+
100 kernels 82 popped 18 unpopped
82100 82%
100
100amount of corn popped
percent yieldamount of kern
percent yie
els in the bag
ld
What you get to eat!
Percent yield
100actual yield
percent yieldtheoretical yield
100actual yield
theoreticalpercent yi
yieldeld
actual yield: the amount obtained in the lab in an actual experiment.
theoretical yield: the expected amount produced if everything reacted completely.
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g 4.87 g
measured experimentally
- There is usually some human error, like not measuring exact amounts carefully
- Maybe the heating time was not long enough; not all the Na2HCO3 reacted
- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too
- CO2 is a gas and does not get measured
Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g 4.87 g
measured experimentally
Let’s calculate the percent yield
100actual yield
percent yieldtheoretical yield
obtained in experiment
calculated
Conversions of Mass to Amounts In Moles
Example
Mass to Mass Calculations
Example
Assignment
• Write a three dollar summary of this section (based on what you learned)
• Answer questions # 4- 6 on page 360• Turn in completed work
• Honors Chemistry Homework– Page 363 # 46 - 57
Vocabulary: 3
• Limiting reactant• Excess reactant
4 slices of bread 4 slices of ham 2 slices of cheese2 ham & cheese
sandwich
Suppose you want to make 2 ham & cheese sandwiches
Recipe:
Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you have4 slices of bread, 4 slices of ham, and 1 slice of cheese?
No, you are limited by the cheese!You can only get 1 ham & cheese sandwich.
Limiting factor
Limiting Reactants
• A limiting reactant is the reactant that limits the amount of the other reactants that can combine to form product in a chemical reaction
• An excess reactant is the reactant that is left over after the other reactant runs out, or the reaction is completed
Excess reactant
For a chemical reaction:
Reactant A is in excess
so the reaction will stop when you
run out of reactant B.
Reactant B is the limiting
reactant.
The amount of product C will
depend on how much reactant B
is present.
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)150 g 60 g
1. What is the limiting reactant?
Step 1: Convert masses to moles
Step 2: Use mole ratios to find the limiting reactant
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)150 g 60 g
1. What is the limiting reactant?
Step 1: Convert masses to moles
Step 2: Use mole ratios to find the limiting reactant
2 32 3 2 3
2 3
1150 0.94
159.7
160.0 2.22
26.98
mole Fe Og Fe O moles Fe O
g Fe O
mole Alg Al moles Al
g Al
2 32 3
2 32 3
20.94 1.88
1
12.22 1.11
2
moles Almoles Fe O moles Al
moles Fe O
mole Fe Omoles Al moles Fe O
moles Al
needed to react with all Fe2O3
needed to react with all Al
available Fe2O3
available Al
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)150 g 60 g
1. What is the limiting reactant?
Step 1: Convert masses to moles
Step 2: Use mole ratios to find the limiting reactant
2 32 3
22
33
1150
159.7
16
0.94
2.0.026.
28
29
molesmole Fe O
g Fe Og Fe O
mole Al
Fe O
g Al mg A
ol s ll
e A
2 3
2 32 3
2 3
20.94
1
12.22
1.88
1.12
1
moles Almoles Al
moles Fe Omoles Fe O
mole Fe Omoles A ml
mololes
es Ae
lF O
needed to react with all Fe2O3
needed to react with all Al
available Fe2O3
available Al
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)150 g 60 g
1. What is the limiting reactant?
2 32 3
22
33
1150
159.7
160.0 2.22
0.
26.98
94 molesmole Fe O
g Fe Og Fe O
mole Al
Fe O
g Al moles Alg Al
2 3
2 32 3
2 3 1.11
20.94 1.88
1
12.22
2
moles Almoles Fe O moles Al
moles Fe O
mole Fe Omoles Al
momole
less Fe O
Al
needed to react with all Al
available Fe2O3
There is not enough Fe2O3 available to react with all the Al, so Fe2O3 is the limiting reactant
Example # 1 & 2
N2(g) + 3H2(g) → 2NH3(s)
The Haber-Bosch process for the synthesis of ammonia:
N2(g) + 3H2(g) → 2NH3(s)
The Haber-Bosch process for the synthesis of ammonia:
Vocabulary: Section 4
• Excess reactant• Percent yield
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
What is left?
2 2
2
1.66 ( ) 1.15 ( )
0.51 ( )
moles N have moles N need
moles N remaining
2 2
28.010.51
114
gmoles N
moleg N
Asked: Amount of excess reactant left
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
28.01 g/mole N2
Solve: 1) How many moles N2 remain?
2) Convert moles to grams
Answer:14 g of N2 will remain at the end of the reaction.
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked: Concentration of CuSO4(aq)
Given: 1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve: 1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of
CuSO4?
42.20 10 moles CuS4
42.20 10 moles CuSO
Have:
Reactants in solution
4
44
42.20 10
2.20 1
1.0
0
moles of solutemolarity
L of solution
moles CuSO
M C
L of w
u
a er
SO
t
Assignment
• On the second section of that sheet of paper, please write six things that you learned from your notes so far that could appear on your test.
Percent Yield• Theoretical yield is the maximum amount of
product that can be produced from a given amount of reactant
• Actual yield is the measured amount of product obtained from a reaction
• Percent yield is the ratio of the actual yield to the theoretical yield
Percent Yield
Percent yield = Actual yield
theoretical yieldX 100 %
Assignment
• Write a three dollar summary of this section (based on what you learned) and using the vocabulary learned this section
• Answer questions # 7- 8 on page 360• Turn in completed work
• Honors Chemistry Homework– Page 364 # 58 - 63
Test:- Next Tuesday or Thursday depending on your class
• Homework requirement: page 360 # 1 - 8
• Make sure you have all work assigned between pages 360 and 363 completed and turned in by your test date.