View
84
Download
0
Category
Tags:
Preview:
DESCRIPTION
Unit 1: Stoichiometry. Chemistry 2202. Stoichiometry. Stoichiometry deals with quantities used in OR produced by a chemical reaction. 3 Parts. Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical Equations (Chp. 4) Solution Stoichiometry (Chp. 6). PART 1 - Mole Calculations. - PowerPoint PPT Presentation
Citation preview
08:09 AM
Unit 1: Stoichiometry
Chemistry 2202
08:09 AM
Stoichiometry
Stoichiometry deals with quantities used in OR produced by a chemical reaction
08:09 AM
3 Parts
Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical
Equations (Chp. 4) Solution Stoichiometry (Chp. 6)
08:09 AM
PART 1 - Mole Calculations
Isotopes and Atomic Mass (pp. 43 - 46)
Avogadro’s number (pp. 47 – 49) Mole Conversions (pp. 50 - 74)
M, MV, NA, n, m, v, N
08:09 AM
Questions
p. 45 #’s 1 – 4p. 46 #’s 1 – 6p. 75 #’s 9 – 12p. 51-53 #’s 5 – 15p. 57 #’s 16 – 19p. 59,60 #’s 20 –
27p. 63,64 #’s 28 - 37
p. 54 #’s 5 – 8p. 65 #’s 2, 4, 5p. 75 #’s 13, 14p. 76 #’s 15, 17–
19,
21-23p. 73 #’s 38 – 43p. 74 #’s 1 – 4p. 76 #’s 26, 27
08:09 AM
PART 1 - Mole Calculations
Percent composition: - given mass (p. 79 - 82) - given the chemical formula (p. 83 - 86)
Empirical Formulas (pp. 87 - 94) Molecular Formulas (pp. 95 - 98) Lab: Formula of a Hydrate
08:09 AM
Questions
p. 82 #’s 1 – 4p. 85 #’s 5 – 8p. 89 #’s 9 – 12p. 91 #’s 13 – 16p. 97 #’s 17 - 20
p. 103 #’s 23 – 24p. 86 #’s 1, 3 – 6p. 94 #’s 1 - 7p. 106 #’s 1 - 3,
6, 7p. 107 – 109
#’s 5 – 23, 25
08:09 AM
Isotopes and Atomic Mass
atomic number - the number of protons in an atom or ion
mass number - the sum of the protons and neutrons in an atom
isotope - atoms which have the same number of protons and electrons but different numbers of neutrons
08:09 AM
Isotopes and Atomic Mass eg.
08:09 AM
Isotopes and Atomic Mass
1735
1737C l C l
08:09 AM
Isotopes and Atomic Mass
1224
1225
1226M g M g M g
not all isotopes are created equal
79 % 10 % 11 %
08:09 AM
Isotopes and Atomic Mass
08:09 AM
Isotopes and Atomic Mass
atomic mass unit (AMU - p.43)- a unit used to describe the mass of
individual atoms- the symbol for the AMU is u- 1 u is 1/12 of the mass of a carbon-12
atom
08:09 AM
Isotopes and Atomic Mass
average atomic mass (AAM)- the AAM is the weighted average of
all the isotopes of an element (p. 45)
p. 14 # 5p. 45 #’s 1 – 4p. 46 #’s 1 – 6 p. 75 #’s 9 - 12
08:09 AM
Finding % Abundance
eg. Br has two naturally occurring isotopes. Br-79 has a mass of 78.92 u and Br-81 has a mass of 80.92 u. If the AAM of Br is 79.90 u, determine the percentage abundance of each isotope.
08:09 AM
Let x = fraction of Br-79Let y = fraction of Br-81
x + y = 178.92x + 80.92y = 79.90
Finding % Abundance
y = 1 - x
08:09 AM
x + y = 178.92x + 80.92y = 79.90
08:09 AM
Avogadro’s Number (p. 47)
p. 48
08:09 AM
Avogadro’s Number The MOLE is a number used by chemists to
count atoms The MOLE is the number of atoms contained in
exactly 12 g of carbon-12. In honor of Amedeo Avogadro, the number of
particles in 1 mol has been called Avogadro’s number.
08:09 AM
How big is Avogadro's number?
An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles.
Avogadro's number of unpopped popcorn kernels spread across the USA, would cover the entire country to a depth of over 9 miles.
08:09 AM
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.
How big is Avogadro's number?
08:09 AM
Avogadro’s Number
1 mole = 6.02214199 x 1023 particles
1 mol = 6.022 x 1023 particles
NA = 6.022 x 1023 particles/mol
08:09 AM
Avogadro’s Number
08:09 AM
Avogadro’s Number
Number of moles Number of atoms
5 mol
0.01 mol
4.65 x 1024 atoms
8.01 x 1021 atoms
7.72 mol
0.0133 mol
6.022 x 1021 atoms
3.011 x 1024 atoms
08:09 AM
Avogadro’s Number
Formulas:
AN
Nn n = # of moles
N = # of particles (atoms, ions, molecules, or formula units)
NA = Avogadro’s #
N = n x NA
08:09 AM
How many moles are contained in the following?
a) 2.56 x 1028 Pb atoms
b) 7.19 x 1021 CO2 molecules
Avogadro’s Number
08:09 AM
Avogadro’s Number
eg. Calculate the number of moles in 4.98 x 1025 atoms of Al.
eg. How many formula units of Na2SO4 are in 5.69 mol of Na2SO4?
# of Na ions? # of Oxygen atoms?
08:09 AM
Avogadro’s Number
1. How many molecules of glucose are in 0.435 mol of C6H12O6?
How many carbon atoms?2. Calculate the number of moles in
a sample of glucose that has 3.56 x 1022 hydrogen atoms.
08:09 AM
Avogadro’s Number
pp. 51 – 53: #’s 5 – 15p. 54: #’s 4 - 8
08:09 AM
Molar Mass
The mass of one mole of a substance is called the molar mass of the substance
eg. 1 mole of Pb has a mass of 207.19 g1 mole of Ag has a mass of 107.87 g
08:09 AM
Molar Mass
The symbol for molar mass is M and the unit is g/mol
eg. MPb = 207.19 g/mol
MAg = 107.87 g/mol
08:09 AM
Molar Mass
The molar mass of a compound is the sum of the molar masses of the elements in the compound
eg. Calculate the molar mass of:a) H2O b) C6H12O6 c) Ca(OH)2
08:09 AM
Molar Mass
H2O has 2 hydrogens and 1 oxygen
2 x 1.01 = 2.021 X 16.00 = 16.00
18.02 g/mol
08:09 AM
Molar Mass
C6H12O6
6 x 12.01 = 72.0612 x 1.01 = 12.126 x 16.00 = 96.00
180.18 g/mol
08:09 AM
Molar Mass
Ca(OH)2
1 x 40.08 = 40.082 x 16.00 = 32.002 x 1.01 = 2.02
74.10 g/mol
Your calculator may not show the zeroes.There should be 2 digits after the decimal when adding molar masses
08:09 AM
Molar Mass
p. 57: #’s 16 – 19 & Molar Masses Handout
1. 151.92 g/mol 7. 58.44 g/mol 2. 120.38 g/mol 8. 100.09 g/mol 3. 286.19 g/mol 9. 44.02 g/mol 4. 100.40 g/mol 10. 248.22 g/mol 5. 74.44 g/mol 11. 115.04 g/mol 6. 78.01 g/mol
08:09 AM
Molar Mass Calculations
AN
Nn
M
mn
N = n x NA
mass
molar
mass
m = n x M
Avogadro’s #
08:09 AM
Molar Mass Calculations
AN
Nn
M
mn
N = n x NA
m = n x M
08:09 AM
Molar Mass Calculations
eg. How many moles are in 25.3 g of NO2?
m = 25.3 gMNO2 = 46.01 g/mol
M
mn
mol/g.
g.
0146
325
mol.5500
08:09 AM
Molar Mass Calculations
eg. What is the mass of 4.69 mol of water?
n = 4.69 molMwater = 18.02 g/mol
m = n x M = (4.69 mol)(18.02 g/mol) = 84.5 g
08:09 AM
Molar Mass Calculations
Practice: p. 59 #’s 20 - 23p. 60 #’s 24 - 27
particles(N)
Moles(n)
Mass(m)
5.98 x 1026 Cu atoms
4.50 g H2O
6.15 mol O3
Particle–Mole-Mass Conversions
08:09 AM
Molar Mass Calculations
Practice: p. 63 #’s 28 - 33p. 64 #’s 34 – 37p. 76 # 15
moles (n)
mass (m)
particles(N)
x M
÷ M÷ NA
x NA
08:09 AM
Molar Mass Calculations
eg. How many molecules are in 26.9 g of water?m = 26.9 gMwater= 18.02 g/mol
NA = 6.022 x 1023 molecules/mol
Find N
08:09 AM
Molar Mass Calculations
M
mn
02.18
9.26
= 1.493 mol H2O
N = n x NA
= 1.493 X 6.022 x 1023
= 8.99 x 1023 molecules
08:09 AM
Molar Mass Calculations
eg. How many molecules are in 4.78 g of glucose?m = 4.78 gMwater= 180.18 g/mol
NA = 6.022 x 1023 molecules/mol
Find N
08:09 AM
Molar Mass Calculations
M
mn
02.18
9.26
= 1.493 mol H2O
N = n x NA
= 1.493 X 6.022 x 1023
= 8.99 x 1023 molecules
08:09 AM
Molar Mass Calculations
eg. A sample of Sn contains 4.69 x 1028 atoms. Calculate its mass.
N = 4.69 x 1028 NA = 6.022 x 1023 molecules/mol
MSn = 118.69 g/mol
Find m
08:09 AM
Molar Mass Calculations
AN
Nn
23
28
10x022.6
10x69.4
= 77,881 mol
m = n x M
= 77881 mol x 118.69 g/mol
= 9.24 x 10 6 g
08:09 AM
Molar Mass Calculations
Practice: p. 63 #’s 28 - 33p. 64 #’s 34 – 37p. 76 # 15
08:09 AM
Molar Mass Calculations
Practice:p. 54 #’s 5 - 8p. 65 #’s 2, 4, 5p. 75 #’s 13, 14, p. 76 #’s 15, 17 – 19, 21 -23
08:09 AM
Molar Volume
•The volume of a gas increases when temperature increases but decreases when pressure increases .
•The volume of gases is measured under conditions of Standard Temperature and Pressure (STP)
08:09 AM
Molar Volume
Standard Pressure – 101.3 kPa Standard Temperature – 0 °C Avogadro hypothesized that equal
volumes of gases at the same temperature and pressure contain equal numbers of molecules.
08:09 AM
Molar Volume
Experimental evidence shows the volume of one mole of ANY GAS at STP is 22.4 L/mol
OR MV = 22.4 L/mol
08:09 AM
Molar Mass Calculations
AN
Nn
M
mn
N = n x NA
m = n x M
MV
vn
v = n x MV
given volume in Litres
08:09 AM
Molar Mass Calculations
moles (n)
mass (m)
particles(N)
x M
÷ M÷ NA
x NA
volume (v)
x MV÷ MV
08:09 AM
Molar Volume
p. 73 #’s 38 – 43
p. 74 #’s 1 – 4
p. 76 #’s 26, 27
08:09 AM
Percent Composition (p. 79)
The mass percent of a compound is the mass of an element in a compound expressed as a percent of the total mass of the compound.
mass percen tm ass o f elem en t
to ta l m ass o f com poundX 100%
08:09 AM
Percent Composition
eg. 8.50 g of a compound was analyzed and found to contain 6.00 g of hydrogen and 2.50 g of carbon. Calculate the mass percent for each element.
p. 82 #’s 1 - 4
08:09 AM
Percent Composition
mass percent may be found using the formula & the molar mass of a compound.
eg. Find the percentage composition for CH4
08:09 AM
Percent Composition
M = 12.01 g/mol + 4(1.01 g/mol) = 12.01 g/mol + 4.04 g/mol
= 16.05 g/mol
% H = (4.04/16.05) X 100% = 25.2 %% C = (12.01/16.05) X 100 % = 74.8 %
p. 85 #’s 5 - 8
p. 86 #’s 1, 3 – 6 p. 107 #’s 6 – 10
08:09 AM
Empirical Formulas
An empirical formula gives the simplest ratio of elements in a compound.A molecular formula shows the actual number of atoms in a molecule of a compound.Ionic compounds are always written as empirical formulas
08:09 AM
Empirical Formulas
Compound Molecular Formula
Empirical Formula
butane C4H10
glucose C6H12O6
water H2O
benzene C6H6
C2H5
CH2O
H2O
CH
08:09 AM
Empirical Formulas (p.87)
08:09 AM
Empirical Formulas
The empirical formula of a compound may be determined by using the % composition of a given compound.
08:09 AM
Empirical Formulas
Method: assume you have 100.0 g of the
compound (ie. change % to g) calculate the moles (n) for each
element divide each n by the smallest n to get
the ratio for the empirical formula
08:09 AM
Empirical Formulas
eg. A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound.
p. 89 #’s 9 - 12
08:09 AM
Empirical Formulas
When finding the EF, the mole ratio may not be a whole number ratio.
eg. A compound contains 84.73% N and 15.27 % H by mass. Determine the empirical formula of the compound.
08:09 AM
p. 90
08:09 AM
Empirical Formulas
eg. A compound contains 89.91% C and 10.08 % H by mass. Determine the empirical formula of the compound.
p. 91 #’s 13 – 16p. 94 #’s 2-4, 6, 7
Answers on p. 109
08:09 AM
MgO Lab
08:09 AM
Molecular Formulas The molecular formula of a compound is
a multiple of the empirical formula.
See p. 95
08:09 AM
Molecular Formulas
To find the molecular formula we need the empirical formula and the molar mass of the compound
eg. The empirical formula of hydrazine is NH2. The molar mass of hydrazine is 32.06 g/mol. What is the molecular formula for hydrazine?
08:09 AM
Molecular Formulas
p. 97 #’s 17 – 20
p. 107, 108 #’s 11 - 14
08:09 AM
CHC analyzer (p. 99 – 101)
1. Describe the operation of a carbon hydrogen combustion analyzer.
2. 22.0 g of carbon dioxide and 10.8 g of water is collected in a CHC analysis. Determine the empirical formula of the hydrocarbon.
p. 101 #’s 21, 22
08:09 AM
Formula of a hydrate
To determine the formula of a hydrate:
- calculate the moles of water
- calculate the moles of anhydrous compound
- determine the simplest ratio
08:09 AM
Formula of a hydrate
eg. Use the data below to determine the value of x in LiCl• xH2O.
mass of crucible = 26.35 g
crucible + hydrate = 42.15 g
crucible + anhydrous compound= 34.94 g
mwater = 42.15 – 34.94 = 7.21 g H2O
mLiCl = 34.94 – 26.35 = 8.59 g LiCl
08:09 AM
eg. Na2CO3. xH2O
crucible = 15.96 g
crucible + hydrate = 22.19 g
crucible + anhydrous compound = 19.67 g
08:09 AM
eg. CoCl2.xH2O
crucible = 151.96 gcrucible + hydrate = 164.35 g crucible + anhydrous compound = 158.23 g
p. 103; # 24 Lab: pp. 104-105
08:09 AM
Formula of a hydrate
mass of empty beaker
mass of beaker & hydrate
mass of beaker & anhydrous compound
08:09 AM
Review – Chp. 3
p. 86 #’s 1, 3 – 6p. 94 #’s 1 – 7p. 103 # 23p. 106 # 7pp. 107–109 #’s 5 – 23, 25
08:09 AM
Test
p. 45 #’s 1 – 4p. 46 #’s 1 – 6p. 75 #’s 9 – 12p. 51-53 #’s 5 – 15p. 57 #’s 16 – 19p. 59,60 #’s 20 –
27p. 63,64 #’s 28 - 37
p. 54 #’s 5 – 8p. 65 #’s 2, 4, 5p. 75 #’s 13, 14p. 76 #’s 15, 17–
19,
21-23p. 73 #’s 38 – 43p. 74 #’s 1 – 4p. 76 #’s 26, 27
08:09 AM
Test
p. 82 #’s 1 – 4p. 85 #’s 5 – 8p. 89 #’s 9 – 12p. 91 #’s 13 – 16p. 97 #’s 17 - 20
p. 103 #’s 23 – 24p. 86 #’s 1, 3 – 6p. 94 #’s 1 - 7p. 106 #’s 1 - 3,
6, 7p. 107 – 109
#’s 5 – 23, 25
08:09 AM
Stoichiometry (Chp.4)
Stoichiometry is the determination of quantities needed for, or produced by, chemical reactions.
Ratios from balanced chemical equations are used to predict quantities.
08:09 AM
Stoichiometry – p. 111
Clubhouse sandwich recipe
08:09 AM
Clubhouse sandwich recipe
Slices of Toast
Slices of Turkey
Strips of Bacon
# of Sandwiche
s
12
27
66
100
Fill in the missing quantities:
08:09 AM
Mole Ratios
A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change.
08:09 AM
Mole Ratios
A mole ratio Come from a balanced chemical
equation Shows the relative amounts of the
reactants/products in moles Is the coefficient for the required species
in the numerator and the coefficient for the given species in the denominator.
08:09 AM
N2(g) + 3 H2 (g) → 2 NH3 (g)
20
66
140
81
08:09 AM
C3H8 + 5 O2 → 3 CO2 + 4 H2O
# #required given xcoeffic ien t
coeffic ien trequired
given
How many moles of CO2 are produced when 31.5 mol of O2 react?
08:09 AM
C3H8 + 5 O2 → 3 CO2 + 4 H2O
# #required given xcoeffic ien t
coeffic ien trequired
given
How many moles of H2O are produced when 1.35 mol of O2 react?
08:09 AM
C3H8 + 5 O2 → 3 CO2 + 4 H2O
# #required given xcoeffic ien t
coeffic ien trequired
given
How many moles of O2 are needed to produce 31.5 mol of CO2?
08:09 AM
C3H8 + 5 O2 → 3 CO2 + 4 H2O
# #required given xcoeffic ien t
coeffic ien trequired
given
How many moles of C3H8 are needed to react with 0.369 mol of O2?
08:09 AM
C5H12 + O2 → CO2 + H2O
# #required given xcoeffic ien t
coeffic ien trequired
given
How many moles of CO2 are produced when 6.35 mol of O2 react?
08:09 AM
Al(s) + Br2(l) → AlBr3(s)
# #required given xcoeffic ien t
coeffic ien trequired
given
How many moles of Br2 are needed to produce 0.315 mol of AlBr3?
p. 115 #’s 4 – 7p. 117 #’s 8 - 10
08:09 AM
Mole to Mole Stoichiometry
1. How many moles of nitrogen gas are needed to produce 6.75 mol of NH3 in a reaction with hydrogen gas?
2. How many moles of silver would be produced if 10.0 mol of silver nitrate reacts with copper metal?
3. How many moles of water are produced when 20.6 mol of CH4 burns?
08:09 AM
Mole to Mole Stoichiometry
1. How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes?
08:09 AM
Mole to Mole Stoichiometry
1. How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes?
08:09 AM
Mass to Mole Stoichiometry
1. How many moles of water are produced when 20.6 g of CH4 burns?
08:09 AM
Mass to Mole Stoichiometry
08:09 AM
Mole to Mass Stoichiometry
1. What mass of CaCl2 is produced when
4.38 mol of Ca(NO3)2 reacts with NaCl?
08:09 AM
08:09 AM
Mole to Mass Stoichiometry
1. What mass of CaCl2 is produced when
4.38 mol of Ca(NO3)2 reacts with NaCl?
1. How many moles of copper would be produced if 20.5 g of copper (II) oxide decomposes?
2. How many moles of water are produced when 5.45 gl of C3H8 burns?
08:09 AM
Mass to Mass Stoichiometry
eg. Calculate the mass of HCl needed to react with 3.56 g of Fe to produce FeCl2.
08:09 AM
Ca(NO3)2 + 2 NaCl → CaCl2 + 2 NaNO3
m ol C aC l2 = 4.38 m ol C a(N O 3)2 x 1 C aC l2
1 C a(N O 3)2
4 .38 m ol C aC l2
08:09 AM
2 CuO → 2 Cu + O2
m ol C u = 0.374 m ol C uO x 2 C u
2 C uO 0 .374 m ol C u
08:09 AM
C3H8 + 5 O2 → 3 CO2 + 4 H2O
m ol H 2O = 5.45 m ol C 3H 8 x 4 H 2O
1 C 3H 8 21.8 m ol H 2O
08:09 AM
Stoichiometry (Chp.4)
Four step stoichiometry:
1. Write a balanced chemical equation
2. Calculate moles given from mass
3. Mole ratio – find moles required
4. Calculate required quantity (mass)
08:09 AM
Stoichiometry (Chp.4)
eg. What mass of CO2 gas is produced when 45.9 g of CH4 burns ?
Step #1
CH4 + 2 O2 → CO2 + 2 H2O
45.9 g ? g
08:09 AM
eg. How many moles of HCl needed to react with 3.56 g of Fe to produce FeCl2.
08:09 AM
Mole Calculations (p. 121 #13)
3.56 g ? g
= 0.06374 mol Fe
Fe + 2 HCl → FeCl2 + H2
55.85g/mol
g3.56nFe Step #2
Step #3
Fe mol 1
HCl mol 2 x Fe mol 0.06374nHCl
= 0.12748 mol HCl
08:09 AM
Mole Calculations
p. 122 #15Given 32.0 g of sulfur (M = 256.56 g/mol) Find mass of ZnS
#2 n = 0.1247 mol S8
#3 n = 0.9976 mol ZnS(M = 97.45 g/mol)
#4 m = 97.2 g ZnS
08:09 AM
Mole Calculations
p. 123 #18Given 33.5 g of H3PO4 (M = 98.00 g/mol) Find mass of MgO
#2 n = 0.3418 mol H3PO4
#3 n = 0.5128 mol MgO(M = 40.31 g/mol)
#4 m = 20.7 g MgO
08:09 AM
Mole Calculations
p. 123 #17Given 25.0 g of Al4C3 (M = 143.95 g/mol) Find volume of CH4
#2 n = 0.174 mol Al4C3
#3 n = 0.522 mol CH4 (MV = 22.4 L/mol)
#4 m = 11.7 L CH4
How many moles of aluminum chloride can be produced from the reaction of chlorine and 10.8 mol of aluminum ?
Cl2(g) + Al(s) → AlCl3(s)
How many moles of magnesium are needed to react with 27 g of iodine to form magnesium iodide?
How many grams of nitrogen are needed to react with 14.0 mol of oxygen to produce nitrogen dioxide ?
N2(g) + O2(g) → NO2(g)
08:09 AM
Mole Calculations (p. 121 #14)
2.34 g ? L
#2 n = 0.05086 mol NO2
#3 n = 0.01272 mol O2
(MV = 22.4 L/mol)
#4 n = (0.01272)(22.4) = 0.285 L O2
08:09 AM
Limiting Reactant (p. 128)
1. 10.0 g of Li requires _______ of Br2
2. 15.0 g of Br2 requires _______ of Li
3. 10.0 g of Li produces _______ of LiBr4. 15.0 g of Br2 produces ______ of LiBr
This problem has _____ g of excess _____and will produce ______ g of LiBr
08:09 AM
Limiting Reactant (p. 128)
10.0 g of Li reacts with 15.0 g of Br2. Calculate the mass of LiBr produced.
08:09 AM
Limiting Reactant (p. 128)
The Limiting Reactant (LR) OR Limiting Reagent (LR) is the substance that is completely used in a chemical reaction.
The Excess Reactant is the reactant that is left over after a reaction is complete.
08:09 AM
Limiting Reactant (p. 128)
eg. 2.00 g of NaI reacts with 2.00 g of Pb(NO3)2. Determine the LR and calculate the amount of PbI2 produced.
write a balanced equation find n for each reactant (Step #2) find moles produced by each
reactant (Step #3)
Pb(NO3)2 + 2 NaI → 2 NaNO3 + PbI2
2.00 g 2.00 g
nPb(NO)3 = 2.00 g 331.21 g/mol = 0.006038 mol
nNaI = 2.00 g 149.89 g/mol = 0.013343 mol
nPbI2 = 0.006038 mol Pb(NO3)2 x 1 mol PbI2
1 mol Pb(NO3)2
= 0.006038 mol PbI2
nPbI2 = 0.013343 mol NaI x 1 mol PbI2
2 mol NaI
= 0.006672 mol PbI2
mPbI2 = 0.006038 mol x 460.99 g/mol
= 2.78 g PbI2
What mass of calcium carbonate will be produced when 20.0 g of calcium phosphate reacts with 15.0 g of sodium carbonate?
(14.2 g)
What mass of barium hydroxide will be produced when 10.0 g of barium nitrate reacts with 30.0 g of sodium hydroxide? (6.56 g)
What volume of hydrogen gas at STP will be produced when 10.0 g of zinc metal reacts with 20.0 g of hydrogen chloride?
Zn + 2 HCl → H2 + ZnCl2
Ba(NO3)2 + 2 NaOH → 2 NaNO3 +Ba(OH)2
08:09 AM
Law of Conservation of Mass (p. 118)
In a chemical reaction, the total mass of reactants always equals the total mass of products.
eg. 2 Na3N → 6 Na + N2
When 500.00 g of Na3N decomposes 323.20 g of N2 is produced. How much Na is produced in this decomposition?
08:09 AM
Law of Conservation of Mass( p. 118)
eg. To produce 90.1 g of water, what mass of hydrogen gas is needed to react with 80.0 g of oxygen?
eg. If 3.55 g of chlorine reacts with exactly2.29 g of sodium, what mass of NaCl willbe produced?
08:09 AM
The theoretical yield is the amount of product that we calculate using stoichiometry
The actual yield is the amount of product obtained from a chemical reaction
Percent yield (p. 137)
08:09 AM
Percent yield (p. 137)
percent y ie ld actua l y ie ld
theoretica l y ie ld x 100
DEMO: silver nitrate + copper
Equation:
Mass AgNO3 =
Mass Cu =
DEMO: silver nitrate + copper
Mass of filter paper and precipitate =
Mass of empty filter paper =
Mass of precipitate =
Recommended