Transcript Testing - Dallas School District

Moles
Moles
Balancing Eqns
Like a recipe:
Reactants
Products
2H2(g) + O2(g)  2H2O(l)
coefficients
subscripts
Moles
Balancing Eqns
Symbols

(s)
(l)
(aq)
(g) or 
Yields or Produces
solid
liquid (pure liquid)
aqueous (dissolved in water)
gas
Moles
Balancing
Copper(II) Chloride reacts with Iron(III)
Sulfate to form Copper(II)Sulfate and
Iron(III) Chloride
Aluminum nitrate reacts with Sodium
hydroxide to form Aluminum hydroxide
and Sodium nitrate
Moles
Some Types of Reactions
1. Synthesis
Al + Cl2  AlCl3
CaO(s) + CO2(g)  CaCO3(s)
2. Decomposition
HgO(s)  Hg(l) + O2(g)
CaCO3(s)  CaO(s) + CO2(g)
Moles
Some Types of Reactions
3. Combustion
Fe + O2  Fe2O3
C4H10 + O2  CO2 + H2O
4. Other Types
Single Replacement
Double Replacement
Moles
Know
your
reactions
well!!!!
Moles
Stoichiometry
Mole
1 dozen = 12 items
1 mole = 6.022 X 1023 atoms/molecules
1 gram hydrogen = 6.022 X 1023 atoms of
hydrogen
Moles
Moles
Moles
Stoichiometry
Grams
Moles
1gH
1 mole H
2gH
12 g C
36 g C
48 g O
Atoms
6.02 X 1023 atms
Moles
Grams
16 g CH4
8 g CH4
88 g CO2
131 g Ba(NO3)2
Stoichiometry
Moles
Molecules
Atoms
Moles
Molar Mass
1. Molar Mass = mass of one mole
2. Element= atomic mass
1 mole of O = 16.0 grams
3. Molecule or Compound – sum of all
the atoms
Moles
Moles
Calculate the molar mass of
Barium
O2
BaCl2?
Fe2(SO4)3?
Molar Mass
Moles
GMA
Grams
Moles
Atoms
1. How many C atoms are present in 18.0 g?
(Ans: 9.03 X 1023 C)
2. What is the mass of 1.20 X 1024 atoms of
Na? (Ans: 45.8 grams)
Moles
GMA
3. What is the mass of 1.51 X 1023 atoms
of Be?(Ans: 2.26 g)
5. How many atoms and grams are in
0.400 mol of Radium? (Ans: 90.4 g,
2.41 X 1023 atoms)
Moles
GMMA
1. Monoatomic Elements (C, Fe, Au)
GMA
2. Molecules and Ionics (H2O, CaCl2, O2)
GMMA
Moles
GMMA
3. Molecules and formula units work the same
when converting
Molecules = Molecular Comps
Formula Units = Ionic Compounds
G
M
M
A
Moles
GMMA
1. How many calcium and chlorine
atoms are in 200.0 grams of Calcium
Chloride?
(Ans: 2.17 X 1024 atoms Cl)
Moles
GMMA
2. How many hydrogen and oxygen
atoms are in 3.60 grams of H2O?
(Ans: 2.41 X 1023 atoms H)
Moles
GMMA
3. Given 3.01X1024 molecules of
SO3, find everything else.
4. Given 3.01 X 1022 molecules of
Iron(III)bromide, find everything
else.
Moles
Mixed Examples
1. How many carbon atoms are in 36.0
grams of carbon (1.81 X 1024)
2. How many carbon atoms are in 36.0
grams of C2H6? (Ans:1.45 X 1024
atoms of C)
Moles
Mixed Examples
Homework Problems (find everything else)
a) 10.0 g C
b) 10.0 g C2H6
c) 4.0 X 1023 atoms of S
d) 4.0 X 1023 molecules of SO2
e) 0.44 moles of SO2
Moles
Empirical Formula
1. Empirical formula - simplest ratio of
the elements in a compound
2. Formula
Empirical Form.
C2H2
Al4S6
C6H12O6
C12H24O12
Moles
Empirical Formula
1. What is the EF of a compound that has
0.900 g Ca and 1.60 g Cl?
Rules
- Go to moles
- Divide by the smaller
Moles
Empirical Formula
2. What is the EF of a compound that is
66.0 % Ca and 34.0% P?
3. What is the EF of a compound that is
43.7 % P and 56.3 % O?
Moles
Molecular Formula
1. Empirical –ratios of the elements
2. Molecular –true number of each
element
Moles
EF
CH2O
Molecular Formula
MF
CH2O (30 g/mol)
C2H4O2 (60 g/mol)
C3H6O3 (90 g/mol)
C4H8O4 (120 g/mol)
Moles
Molecular Formula
1. What is the MF of benzene if it has an
EF of CH and a molar mass of 78.0 g?
2. What is the MF of a compound that is
40.9% C, 4.58 % H and 54.5 % O? It
has a molar mass between 350 and 360
g/mol.
Moles
What coefficients mean:
2 Na +
Cl2 
2 Na
1 Cl2
4 Na
6 Na
Reaction Stoich.
2NaCl
2NaCl
Moles
Cl2 
4 Cl2
2 moles Na
10 moles Na
2 Na +
Reaction Stoich.
2NaCl
ONLY WORKS FOR MOLES AND MOLECULES
Moles
Reaction Stoich.
1. How many moles of H2 and O2
must react to form 6 moles of
H2O?
2. How many moles of KCl and O2 are
formed from the decomposition of 6
moles of KClO3?
Moles
Reaction Stoich.
3. How many grams of oxygen are
needed to react with 14.6 g of Na to
form Na2O? (Ans: 5.08 g)
4. How many grams of P4 and O2 are
needed to make 3.62 g of P2O5?
(Ans: 1.58 g, 2.04 g)
Moles
Reaction Stoich.
5. What mass of oxygen is needed to
react with 16.7 g of iron to form
Iron(III)oxide? (Ans: 7.18 g)
4Fe + 3O2  2Fe2O3
Moles
Calculate the mass of sodium bromide and oxygen
that are formed from the decomposition of 50.0
grams of sodium bromate (NaBrO3).
(34.1 g NaBr, 15.9 g O2)
Moles
Limiting Reactant
1. Sandwich analogy:
13 slices of bread
4 pieces of turkey
Maximum # of sandwiches?
2. Limiting Reactant – Totally consumed in a
reaction. No leftovers
Moles
Limiting Reactant
1. How many grams of H2SO4 can be formed
from the rxn of 5.00 moles of SO3 and
2.00 moles of H2O?
SO3 + H2O  H2SO4
(Ans: 196 g)
Moles
Limiting Reactant
1. How many grams of H2O can be formed
from the rxn of 6.00 moles of H2 and 4.00
moles of O2?
O2 + H2  H2O
Moles
Limiting Reactant
2. How many grams of NaCl can be formed
from the reaction of 0.300 mol of Na and
0.100 mol of Cl2?
2Na + Cl2  2NaCl
(Ans: 11.7 g)
Moles
Limiting Reactant
3. How many grams of Ag can be formed
from the rxn of 2.00 g of Zn and 2.50 g of
silver nitrate? How much excess reactant
remains?
Zn + AgNO3  Ag + Zn(NO3)2
(Ans: 1.59 g Ag, 1.52 g xs zinc)
Moles
Limiting Reactant
4. How many grams of Ba3(PO4)2 can be formed
from the rxn of 3.50 g of Na3PO4 and 6.40 g
of Ba(NO3)2?
Na3PO4 + Ba(NO3)2  Ba3(PO4)2 + NaNO3
(Ans: 4.92 g)
Moles
Limiting Reactant
6. How many grams of Ag2S can be formed
from the rxn of 15.6 g of Ag and 2.97 g of
H2S? (Assume O2 is in excess)
4Ag + 2H2S + O2  2Ag2S + 2H2O
(Ans: 18.1 g)
Moles
Percent Yield
A. Formula:
Actual Yield
X 100 = % Yield
Theoretical Yield
Moles
Percent Yield
1. What is the % yield if you start with 10.00
grams of C and obtain 1.49 g of H2 gas?
C + H2O  CO + H2
(Ans: 89.4%)
Moles
Percent Yield
2. Carbon was heated strongly in sulfur(S8)
to form carbon disulfide. What is the
percent yield if you start with 13.51 g of
sulfur and collect 12.5 g of CS2?
4C + S8  4CS2
(Ans: 78.0%)
Moles
Percent Yield
3. 36.7 grams of CO2 were formed from the rxn
of 40.0 g of CH3OH and 46.0 g of O2. What
is the % yield?
2CH3OH +
(ANS: 87%)
3O2  2CO2 + 4H2O
Moles
Moles
In this experiment, magnesium chloride was
prepared and its empirical formula was compared
to the accepted formula of MgCl2. To prepare
magnesium chloride, 0.40 grams of magnesium
powder was combined with 10 mL of 0.10 M
HCl. The mixture was allowed to react, and
heated to dryness. The mass of the resulting
crystals was used to calculate the empirical
formula. The average calculated formula of
MgCl1.8 had a 10% error and a range was 0.40
chlorine atoms. This procedure was not effective
because while it was accurate, it was not precise.
Moles
8a) SO3 + H2O  H2SO4
b) B2S3 + 6H2O  2H3BO3 + 3H2S
c) 4PH3 + 8O2  6H2O + P4O10
d) 2Hg(NO3)2  2HgO + 4NO2 + O2
e) Cu + 2H2SO4  CuSO4 + SO2 + 2H2O
Moles
12.a)
b)
c)
d)
e)
f)
g)
6
1
2
1
3
2
4
1
3
2
6
2
1
9
2
2
1
3
1
1
4
4
2
6
2
10 2
14. a)
b)
c)
d)
e)
1
1
1
2
1
1
6
2
2
2
1
2
2
4
1
3
1
1
1 2
18.a) 4Al + 3O
Moles
 2Al2O3
b) Cu(OH)2  CuO + H2O
c) C7H16 + 11O2  7CO2 + 8H2O
d) 2C5H12O + 15O2  10CO2 + 12H2O
20. 2 9 6 6
1 1 2
1 6 5 3
1 3 2
1 1 2
22.a) 44.0 g/mol
b) 122.0 g/mol
c) 58.3 g/mol
d) 60.0 g/mol e) 130.0 amu
2
24. 26.0 g/mole
Moles
176.0 g/mole
132.1 g/mole
300.1 g/mole
272.0 g/mole
305.0 g/mole
92.3% C
4.5% H
6.1% H
65.01% Pt
11.8% O
70.8 % C
Moles
46 a)K3PO4
b) Na2SiF6
48 a) H2C2O4
b) C4H8O2
50 a) C13H18O2 b) C5H14N2
c) C12H12N2O3
c) C9H13O3N
Moles
58.a) 0.800 mol CO2 b) 14.7 g C6H12O6
c) 7.16 g CO2
60.a) 0.939 mol Fe2O3
c) 105 g Fe
b) 78.9 g CO
d) 229 g= 229 g
48.a) H C O
Moles
b) C4H8O2
50. a) C13H8O2 b)C5H14N2
c) C9H13O3N
58.a) 0.800 mol CO2
b) 14.7 g C6H12O6
c) 7.18 g CO2
60.a) Fe2O3 + 3CO  2Fe + 3CO2
b) 78.9 g CO2
c) 124 g CO2
d) 229 g = 229 g
2 2
4
Moles
62.a) CaH2 + 2H2O  Ca(OH)2 + 2H2
b) 88.75 g CaH2
64. a) 15.6 mol O2 b) 35.0 g O2
c) 9175.1 g
72.0.167 mol Al2(SO4)3 form
0.333 mol Al(OH)3 react
0.167 mol AL(OH)3 remain
74.a) O2 is limiting reactant
b) 1.86 g H2O produced
c) 0.329 g NH3 remain
d) 4.25 g = 4.25 g
Moles
76. 5.24 g H2SO4
6.99 g PbSO4
2.77 g HC2H3O2
78. C2H6 + Cl2  C2H5Cl + HCl
232 g C2H5Cl (theoretical yield)
88.8% yield
80. Actual yield of Na2S = 1.80 g (1.95 g is the
theoretical yield)
Moles
The atmosphere of Jupiter is composed almost
entirely of hydrogen (H2) and helium (He). If the
average molar mass of Jupiter’s atmosphere is
2.254 g/mole, calculate the percent composition.
Moles