Download Quantity relationships: How much

SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P1
Quantity relationships: How much? How many?
How many dragonflies’ wings are required for 4 dragonflies to
be able to fly?
1 Dragonfly
→
4 wings
0
0
?
SCC-CH110/UCD-CH41C
Same Units
Chapter: 9
Instructor: J.T.
P2
1 DF
4 wings
=
5 DFs
?
Same Units
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P3
Quantity Relationships in Chemical Reactions:
How to solve stoichiometry problems:
What we need to know are two subjects as:
1) A balanced chemical reaction.
2) Relationship between mass and mole.
EXAMPLE:
How many moles of oxygen are required to burn 2.40
moles of ethane gas?
2 C2H6 (g) + 7 O2 (g)
Given
→
4 CO2 (g) + 6 H2O (l)
Wanted
2 mol
C2H6
2.40 mol
C 2H 6
0
7 mol
O2
?
0
SCC-CH110/UCD-CH41C
Chapter: 9
2 mol
2.4 mol
Instructor: J.T.
P4
7 mol
?
=
2 mol (C2H6) × ? = 7 mol (O2) × 2.4 mol (C2H6)
7 mol
?=
× 2.4 mol
2 mol
***********************************************
EXAMPLE: Calculate the number of grams of oxygen that
are required to burn 155 g of ethane.
2 C2H6 (g) + 7 O2 (g)
→
4 CO2 (g) + 6 H2O (l)
Given
2 mol
C 2H 6
155 g
g
30.06 (
)
mol
155 g
C 2H 6
Same Units
Wanted
0
7 mol
O2
?
0
Same Units
SCC-CH110/UCD-CH41C
Chapter: 9
2 mol
5.15 mol
Same Units
Instructor: J.T.
=
7 mol
?
Same Units
7
? = 5.15 ×
2
? = 18.03 mol (oxygen)
Wanted in gram!
mass
mole =
molar mass
molar mass of oxygen = 2 × 16.00 (g/mol)
? = 18.03 mol × 32.00 (g/mol)
? = 577 g O2
P5
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P6
EXAMPLE:
How many milligrams of Nickel Chloride are in a
solution if 503 mg of Silver Chloride is precipitated in the
reaction of Silver Nitrate and Nickel Chloride solution?
2 AgNO3 (aq) + NiCl2 (aq) → 2 AgCl (s) + Ni(NO3)2 (aq)
Wanted
1 mol
NiCl2
Given
2 mol
0
?
0
AgCl
503 mg
mg (AgCl)
AgCl
mol (AgCl)
?
Same Units
¾ (mg) to (g):
503 mg × (1 g/ 1000 mg) = 0.503 g
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P7
¾ (mass) to (mole)
mole = mass / molar mass
mole of AgCl = 0.503 g ÷ 143.35 (g/mol)
mole of AgCl = 3.51 × 10-3 mol
1 mol
2 mol
=
?
72.1 mol
? = 3.51 × 10 −3 mol AgCl ×
1 mol NiCl2
2 mol AgCl
? = 1.75 × 10-3 mol NiCl2
Wanted in milligram!
mass
mole =
molar mass
molar mass of NiCl2= 129.59 (g/mol)
Mass of NiCl2 = 1.75 × 10-3 mol × 129.59 (g/mol) =0.227 g
Mass of NiCl2 = 227 mg
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P8
Gas Stoichiometry at Standard Temperature and Pressure:
One mole of any gas at STP occupies 22.4 L
EXAMPLE:
What volume of hydrogen, at STP, can be released by
42.7 g of zinc as it reacts with hydrochloric acid?
Zn(s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq)
Given
1 mol
Zn
42.7 g
g
65.39 (
)
mol
Wanted
22.4 L
0
?
42.7 g
0
Same Units
1 mol
0.653 mol
=
22.4 L
?
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P9
EXAMPLE:
1.30L of gaseous ethylene is burned. What volume of
oxygen is required if both gas volume are measured at STP?
→ 2CO2 (g) + 2 H2O(l)
C2H4 (g) + 3O2 (g)
Given
Wanted
1 mol
C 2H 4
22.4 L
3 mol
O2
0
3 × 22.4 L
0
?
1.30 L
22.4 L
1.30 L
0
=
67.2
?
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P10
Gas Stoichiometry at nonstandard condition:
In STP condition: T= 273 K, and P =760 torr
P1 V1
T1
=
STP =
P2 V2
T2
nonstandard
In the previous example for the reaction:
Zn (s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq)
We calculated the volume of hydrogen as 14.6 L, now let’s
change the problem. Instead of measuring the hydrogen
volume at STP, let’s use t = 21 °C and P = 748 torr.
TK=21+273 =294
760 × 14.6 748 V2
=
2731
294
STP
V2= 16.0 L
=
nonstandard
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P11
14 transistors
22 resistors
18 capacitors
How many of the above electronic circuits (ec) can you
assemble from 22 resistors, 18 capacitors, and 14 transistors?
2
+4
→
+3
Calculate the theoretical yield of ec:
1 ec
14 T ×
2T
= 7 ec
1 ec
22 R ×
= 5.5 ec ≈ 5 ec
4T
1 ec
18 C ×
= 6 ec
3C
Resistor is the limiting item, because 22 of resistors give 5 ec.
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P12
Limiting Reactants:
Example:
The fertilizer ammonium nitrate can be made by direct
combination of ammonia with nitric acid. If 74.4 g of
ammonia is reacted with 159 g of nitric acid, how many
grams of ammonium nitrate can be produced? Also calculate
the mass of unreacted reactant that is in excess.
NH3 + HNO3
→
NH4NO3
Convert the number of grams of each reactant to moles.
NH3:
74.4 g
= 4.37 mol
g
17.03
mol
159 g
HNO3: 63.02 g = 2.52 mol
mol
1 mol NH 4 NO3
4.37 mol NH 3 ×
= 4.37 mol NH 4 NO3
1 mol NH 3
1 mol NH 4 NO3
2.52 mol HNO3 ×
= 2.52 mol NH 4 NO3
1 mol HNO3
∴ The reactant that yields the smaller amount of product is limiting reactant.
1 mol of NH3 ≡ 1 mol of HNO3
2.52 mol
≡ ?
? =2.52 mol of HNO3
4.37 mol – 2.52 mol = 1.85 mol NH3 left
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P13
1.85 × 17.03 = 31.51 g NH3 left
A mixture of 5.0 g of H2 (g) and 10.0 g of O2(g) is
ignited. Water forms according to the following
combination reaction:
2H2(g) +O2(g) → 2H2O(g)
Which reactant is limiting? How much water will the
reaction produce?
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P14
Percent Yield:
The calculated amount of product if is based on the
assumption that all of the reactant is converted into product is
called the theoretical yield. In laboratory or in industrial
production, the actual amount of product isolated from a
reaction is usually less than the theoretical yield, and it is
called actual yield.
General solution:
αA + βB
→
γC
Given: n gram of A
Given: actual yield
Wanted: %yield
Step 1:
Convert mass of A into the mole of A.
Step 2:
Calculate the theoretical yield:
γ mol C
Number of moles A ×
α mol A
Step 3:
Convert mole of C into the mass of C.
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P15
Step 4:
actual yield
%Yield =
× 100
theoretical yield
Example:
A general chemistry student, preparing copper metal by
the reaction of 1.274 g of copper sulfate with zinc metal,
obtained a yield of 0.392 g copper. What was the percent
yield?
To calculate %yield, we need both the theoretical yield of
copper and the actual yield.
CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq)
Mass to Mole:
1.274 g CuSO 4
= 7.982 × 10 −3 mol CuSO 4
159.6 g
mol
The theoretical yield:
7.982 × 10 −3 mol CuSO 4 ×
1 mol Cu
1 mol CuSO 4
= 7.982 × 10 −3 mol Cu
Mole to Mass:
7.982 × 10 −3 mol Cu × 63.55 g
% Yield
=
mol
= 0.5072 g Cu
actual yield
0.392 g
× 100 =
× 100 = 77.3 %
theoretical yield
0.5072 g
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
Liquid Oxygen
Liquid Hydrogen
Main Engines
P16
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P17
Thermochemical Equations:
H2 (l) + O2 (l) → H2O (g) + energy
System
Gives off heat
2 NH3 (g) + energy
System
Exothermic -ΔH
→ N2(g) + 3 H2 (g)
Absorbs heat
Endothermic +ΔH
The SI unit for energy: joule (J)
How many kilojoule of energy are released when 73.0 g
C2H6 burns according to:
2C2H6 (g) +7 O2 (g) → 4 CO2(g) + 6H2O (l)+ 3119KJ
2 mol
C 2H 6
0
3119
KJ
?
73.0 g
g
30.05 (
)
mol
73.0 g
0
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P18
ΔH =2.82 × 103 KJ for the photosynthesis reaction by which
plants use energy from the sun to form one mole sugar from
carbon dioxide and water. How much energy is required to
form 454g (1lb) of simple sugar C6H12 O6?
6 CO2 (g) +6 H2O (g) +
→ C6H12O6 (s) + 6O2 (g)
1 mol
Suga
0
2.82 × 103
KJ
?
454 g
g
180 (
)
mol
454 g
0
∴ 7.11 × 103 KJ
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P19
SCC-CH110/UCD-CH41C
Chapter: 9
Instructor: J.T.
P20