Chapter 12 “Stoichiometry” Chemistry Tutorial Stoichiometry Mr. Mole.
is the study of the relative quantities stoichiometry A. Using Mole Ratios Stoichiometry and...
-
Upload
yasmin-schroeder -
Category
Documents
-
view
219 -
download
3
Transcript of is the study of the relative quantities stoichiometry A. Using Mole Ratios Stoichiometry and...
is the study of the relative quantitiesstoichiometry
A. Using Mole Ratios Stoichiometry and Quantitative Analysis
of reactants and products in achemical reaction – stoich video
you can use the number of moles for a given reactant or product to find the moles for any other reactant or product
The equation I use is the following:
WR = Wn
GR = Gn
Example
Consider the following chemical reaction:
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)
a) Write the ratio for all components of the reaction.
b) What amount, in moles, of CO2(g) is formed if 2.50 mol of C2H6(g) reacts?
2:7:4:6
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)n = 2.50 mol n = 2.50 mol
= 5.00 mol
42
c) What amount, in moles, of O2(g) is required to react with 10.2 mol of C2H6(g)?
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)n = 10.2 mol n = 10.2
mol = 35.7 mol
72
d) What amount, in moles, of H2O(g) is formed when 100 mmol of CO2(g) is formed?
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)
n = 100 mmol n = 100 mmol
= 150 mmol
64
= 0.150 mol
gravimetric =mass measurements
B. Gravimetric Stoichiometry
Steps
4. Calculate of the wanted species using
3. Find the of the species using
2. Find the of the species using
Write a including the states. Write the information
balanced equationgiven.
moles givenn=m M
moles wantedmole ratio
mass
m=nM
WR = Wn
GR = Gn
Example 1 Iron is produced by the reaction of iron(III) oxide with carbon monoxide to produce iron and carbon dioxide. What mass of iron(III) oxide is required to produce 1000 g of iron?
CO2(g) m=?
M = 159.70 g/mol Step 1: n = m
M = 1000 g 55.85g/mol = 17.90… mol
Step 3 : m = nM = (8.95… mol)(159.70 g/mol) = 1429.7225 g = 1430 g
Fe2O3(s) CO(g) Fe(s) ++ 3 2 31
12
Step 2 : n = 17.90… x= 8.95… mol
m = 1000 g M = 55.85 g/mol
w g
Example 2 The decomposition of the mineral malachite, Cu2(CO3)(OH)2(s), yields copper(II) oxide, carbon dioxide and water vapour. What mass of copper(II) oxide is produced from 1.00 g of malachite?
H2O(g)
m=?M = 79.55 g/mol
Step 3: m = nM = (0.00904… mol)(79.55 g/mol) = 0.7194862 g = 0.719 g
Step 1: n = m M = 1.00 g 221.13 g/mol = 0.00452… mol
Cu2(CO3)(OH)2(s) CuO(s) +CO2(g)+2 1 11
2 1
Step 2: n = 0.00452… x
m = 1.00 g M = 221.13 g/mol
= 0.00904… mol
g w
use to perform calculations
concentrationC. Solution Stoichiometry
c = n v
Example 1 What volume of 14.8 mol/L KOH is needed to react completely with 1.50 L of 12.9 mol/L sulphuric acid?
v=?c= 14.8 mol/L
STEP 2: n = 19.35 x 2
1 = 38.70 mol
v = 1.50 L c = 12.9 mol/L
H2O(l)KOH(aq) H2SO4(aq) K2SO4(aq) ++ 1 1 22
Step 1: n = cv = 12.9 mol/L x 1.50 L = 19.35 mol Step 3: v = n
c = 38.70 mol 14.8 mol/L = 2.6148649 L = 2.61 L
gw
you can use the Law of Combining Volumes when the pressure and temperature conditions are constant for both the
D. Law of Combining Volumes
reactants and the products
Example Consider the following reaction:
N2(g) + 2 O2(g) N2O4(g)
a) What is the mole ratio for O2(g) and N2O4(g)?
b) What is the volume ratio for O2(g) and N2O4(g)?
c) If 1 mol of N2O4(g) is produced, how many moles of O2(g) must be consumed?
2:1
2:1
1 mol of N2O4(g) 2 = 2 mol of O2(g) 1
d) If 1 L of N2O4(g) is produced, what volume of O2(g) must be consumed?
e) If 2.5 L of N2(g) is consumed, what volume of O2(g) must be consumed?
1 L of N2O4(g) 2 = 2 L of O2(g) 1
2.5 L of N2(g) 2 = 5.0 L of O2(g) 1
if the other information is given for the chemicals in the reaction (eg. mass), use to perform calculations
E. Gas Stoichiometry
ideal gas law
PV = nRT
Example 1 If 300 g of propane burns in a gas barbeque, what volume of oxygen at SATP is required for the reaction?
v=? P = 100.000 kPaT = 298.15 KR = 8.314 kPaL/molK
m = 300 g M = 44.11 g/mol
n = 6.80… mol x 5 1
= 34.0… mol
H2O(g) C3H8(g) O2(g) CO2(g)++ 5 3 41
PV = nRT(100.000kPa)V = (34.0… mol)(8.314)(298.15K) V = 842.94…L = 843 L
n = m M = 300 g 44.11 g/mol = 6.80… mol
g w
limiting reagent = the that determines in a reaction
let’s make double burgers…
1 bun + 2 meat patties 1 double burger2 buns + 4 meat patties 2 buns + 2 meat patties
2 buns + one million meat patties
excess reagent = the that is present in than necessary
21
2excess
excess
limiting
limiting
Chapter 8: Applications of Stoichiometry 8.1 Limiting and Excess Reagents
double burgersdouble burger
double burgers
reactanthow much product can be formed
reactantlarger quantities
Steps
1.
2.
3.
Write the including states.
Calculate the of using
Use to calculate the answer.
balanced chemical equation,
number of moles each reactantn=m or C = n/v M
limiting reagent moles
Do step 2 for both reactants. WR = Wn
GR = Gn
- Use the lower n for step 3. (it’s the limiting reagent)
Example 1 When 80.0 g copper and 25.0 g of sulphur react, which reactant is limiting and what is the maximum amount of copper(I) sulphide that can be produced?
16 Cu(s) + 1 S8(s) 8Cu2S(s)
m=? M = 159.17 g/mol
m = 80.0 g M = 63.55 g/mol
n =
m = 25.0 g M = 256.56 g/mol
n/16 = 0.0786…mol n/1 = 0.0974… mol\ limiting \ excess
1.25…mol
= 0.629… mol
n = 80.0 g 63.55g/mol = 1.25… mol
n = 25.0 g 256.56g/mol = 0.0974… mol
8/16
m = (0.629…mol ) (159.17 g/mol) = 100.17… g = 100 g
g w
Example 2You are supplied with 9.00 g of KCl and 6.50 g of AgNO3. What is the mass of the precipitate formed when these two chemicals react?
m = ?M = 143.32 g/mol
m = 9.00 g M = 74.55 g/mol
n =
m = 6.50 g M = 169.88 g/mol
n/1 = 0.120…mol n/1 = 0.0382… mol\ excess \ limiting
0.0382… mol
n = 9.00 g 74.55g/mol =0.120… mol
n = 6.50 g 169.88g/mol = 0.0382… mol
1/1
1 KCl(aq) + 1 AgNO3(aq) ® 1 KNO3(aq) + 1 AgCl(s)
m = (0.0382…mol ) (143.32 g/mol) = 5.48 g
Example 3A 200 mL sample of a 0.221 mol/L mercury (II) chloride solution reacts with 100.0 mL of a 0.500 mol/L solution of sodium sulphide. What is the mass of the precipitate formed?
m = ? M = 232.66 g/mol
C = 0.221 mol/L V = 0.200 L
n =
C = 0.500 mol/L V = 0.1000 L
n/1 = 0.0442…mol n/1 = 0.0500… mol\ limiting
0.0442…mol
m = (0.0442…mol ) (232.66 g/mol) =10.3 g
n = CV=(0.221 mol/L) (0.200 L) =0.0442… mol
1/1
1 HgCl2(aq) + 1 Na2S(aq) ® 2 NaCl(aq) + 1 HgS(s)
n = CV=(0.500 mol/L) (0.1000 L) =0.0500… mol
the is called the
the quantity of the product is called the
it is for the predicted and experimental yield to be
predicted or theoretical yieldexpected amount of product
experimental or actual yieldactually obtained
the sameextremely rare
8.2 Predicted and Experimental Yield
factors affecting the experimental yield include: 1. two chemicals can react to give
…called
eg) C(s) and O2(g) can react to form CO2(g) or CO(g)
2. reaction is very
3. methods
4. reactant or product
5. reaction doesn’t go to
different productscompeting reactions
slow
collection and transfer
purity
completion
ideally, percent yield should be as close to as possible
100%
percent yield is calculated as follows:
% yield = actual yield 100 predicted yield
we can also calculate our for the experiment, which tells us
the closer to the percent error is, the
% error = actual yield – predicted yield 100 predicted yield
how far we are from the theoretical yield
% error
better the experiment
zero
Example 1 Calculate the % error and % yield for the following:
predicted mass of ppt = 6.20 gactual mass of ppt = 7.12 g
% error = 7.12 g - 6.20 g x 100 6.20 g
= 14.8 %
% yield = 7.12 g x 100 6.20 g = 115 %
% error = 93.5 g - 100 g x 100 100 g
= -6.50 %
% yield = 93.5 g x 100 100 g = 93.5 %
Example 2 Calculate the % error and % yield for the following:
predicted mass of ppt = 100 gactual mass of ppt = 93.5 g
in solution stoichiometry, sometimes you don’t have enough information to solve the problem on paper
A. Titrations
eg) 10 mL of acetic acid reacts with a 0.202 mol/L NaOH solution. What is the concentration of the acetic acid?
CH3COOH(aq) + NaOH(aq) H2O(l) + NaCH3COO(aq)x mol/L
v = 0.0100 L
c = 0.202 mol/Lv = ??? *****
**** you need this volume in order to solve the problem
8.3 Acid-Base Titration
a is a used to find the of substances so you can calculate
is when a solution of concentration, a , is reacted with a solution of concentration
both need to be since their concentrations will
titration procedure
concentration
standardization knownstandard solution
unknown
strong acids and strong basesstandardized
change over time
volume
a solution called the is transferred from a precisely marked tube called a to a containing the and anindicator
sampleflaskburette
titrant
an indicator (eg. methyl orange, bromothymol blue) is used because a sudden change in colourindicates the
completion of the reaction
the endpoint is the point where the titrant reacts completely with the sample
equivalence point is the volume needed to reach the endpoint
you need a minimum of 3 trials within of each other to ensure results are accurate
0.20 mL
Here is a website to view a virtual titration:
Virtual Titration
Practice using this simulation.
ExampleA 10.00 ml sample of HCl(aq) was titrated with a standardized solution of 0.685 mol/L NaOH(aq). Bromothymol blue indicator was used and it changes from yellow to blue at the endpoint. What is the concentration of the HCl(aq)? Note: HCl(aq) “is titrated with” NaOH(aq)
sample in flask
titrant in burette
Trial
1 overshoo
t
2
3
4
Final Volume (mL)
Initial Volume (mL)
Volume NaOH (mL)
Endpoint Colour
Data Table
average volume =
NaCl(aq) x mol/LV = 10.00 mL = 0.01000 L n = C = n V = =
C = 0.685 mol/L V = mL = L n = = mol
( + + )
3= mL
HCl(aq) + NaOH(aq)→ H2O(l) +1 1 1 1
B. Titration Curvesa plot of the
is called a pH titration curve pH vs. the volume of titrant added
titration curves areS-shaped
when a is titrated with a the will always have a pH of 7 (at 25C)
equivalence pointstrong monoprotic base,strong monoprotic acid
the on the curve is always the pH of the sample
first point
the pH changes at first…this is called thebuffer region
very gradually
as the endpoint is approached, the pH changesvery rapidly
the is (goes to infinity/never ends) to the pH of thetitrant
asymptoticovertitration
Strong Acid Titrated with Strong Base
pH
volume of titrant added (mL)
7
0
14
Strong Base Titrated with Strong Acid
pH
volume of titrant added (mL)
7
0
14
C. Indicators for Titrations can be used to carry out a
titration but it is much more convenient to use an
the indicator should change colour
the pH of the endpoint should
indicatorpH meters
immediately after the endpoint is reached
fall within the pH range of the indicator
Review assignment: p. 328 #1-33 (omit 22)