Transcript Chapter 12 Powerpoint

WARM UP

What's the mass of 1 mole of water?

What's the mass of 2 moles of Carbon Dioxide?

What’s the mass of 3.5 moles of Sodium Chloride?
CHAPTER 12
STOICHIOMETRY
WHAT IS IT??

Chemists use balanced chemical equations as a
basis to calculate
How much reactant is needed
B. How much product is formed in a reaction.
A.

The calculation of quantities in chemical
reactions is called Stoichiometry
stoicheion "element“
 metron "measure"

LIKE A RECIPE

A balanced chemical equation provides the same kind of
quantitative information that a recipe does.
LET’S LOOK AT AN EVERYDAY EXAMPLE!

In a five day work
week, tiny tike is
scheduled to make
640 tricycles. How
many wheels should
be in the plant on
Monday morning to
make those tricycles?
INTERPRETING CHEMICAL EQUATIONS
A
balanced chemical equation can be interpreted in terms of
different quantities, including numbers of
 Representative
Particles

Atoms

Molecules

Functional Units
 Moles
 Mass
- Grams
 Volume
- Liters
INTERPRETING CHEMICAL EQUATIONS
This reaction requires the use of
a catalyst, high pressure (100–
1,000 atmospheres), and
elevated temperature (400–550
°C [750–1020 °F]).
USES OF AMMONIA
The major use of ammonia is as a fertilizer
 Explosives (e.g., Trinitrotoluene (TNT),
nitroglycerin, and nitrocellulose)
 Synthetic fibers, such as nylon and rayon
 Dyeing and scouring of cotton, wool, and silk
 A coolant in refrigeration and air-conditioning
equipment
 Minor uses is inclusion in certain household
cleansing agents

INTERPRETING (CONT.)
INTERPRETING (CONT.)

Always remember : Mass and atoms are always
conserved in a chemical reaction.
EXAMPLE OF INTERPRETING A CHEMICAL
REACTION
SECTION 2
CHEMICAL CALCULATIONS
WRITING AND USING MOLE RATIOS

Mole Ratios are used to convert between:
moles of reactant and moles of product
 moles of reactants
 moles of products


A mole ratio is a conversion factor derived from the
coefficients of a balanced chemical equation
interpreted in terms of moles.
EXAMPLE #1
EXAMPLE #2
PRACTICE ON YOUR OWN

Page 360 #11
11. This equation shows the formation of aluminum oxide, which
is found on the surface of aluminum objects exposed to the air.
4Al(s) + 3O2(g)  2Al2O3(s)
a. Write the six mole ratios that can be derived from this
equation.
b. How many moles of aluminum are needed to form 3.7 mol
Al2O3?
MASS TO MASS CALCULATIONS
EXAMPLE #4
PRACTICE ON YOUR OWN
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Page 361 # 13
13. Acetylene gas (C2H2) is produced by adding water to calcium
carbide (CaC2).
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
How many grams of acetylene are produced by adding water to 5.00 g
CaC2?
PRACTICE ON YOUR OWN

Page 361 # 13
13. Acetylene gas (C2H2) is produced by adding water to calcium
carbide (CaC2).
CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq)
How many grams of acetylene are produced by adding water to 5.00 g
CaC2?
INTERPRETING CHEMICAL EQUATIONS
A
balanced chemical equation can be interpreted in terms of
different quantities, including numbers of
 Moles
 Mass
- Grams
 Representative
Particles

Atoms

Molecules

Functional Units
 Volume
- Liters
INTERPRETING (CONT.)
OTHER STOICHIOMETRIC CALCULATIONS
 In
a typical stoichiometric problem,
 given quantity is first converted to moles
 mole ratio from the balanced equation is
used to calculate the number of moles of
the wanted substance
unit
unit
 moles are converted to any other unit of
given
measurement
relatedrequested
to the unit mole,
as the problem requires
PROBLEM SOLVING APPROACH
PROBLEM SOLVING APPROACH
unit
given
unit
requested
Analyze List the knowns and the unknown.
Knowns
• mass of water 29.2 g H2O
• 2 mol H2O = 1 mol O2 (from balanced equation)
• 1 mol H2O = 18.0 g H2O (molar mass)
• 1 mol O2 = 6.02 x 1023 molecules O2
Unknown
• molecules of oxygen = ? molecules O2
The following calculations need to be done:
g H2O  mol H2O  mol O2  molecules O2
The appropriate mole ratio relating mol O2 to mol H2O from the
balanced equation is 1 mole O2 /2 mol H2O.
PRACTICE ON YOUR OWN

Page 364 # 15
15. How many molecules of oxygen are produced by the
decomposition of 6.54 g of potassium chlorate (KClO3)?
2KClO3(s)  2KCl(s) + 3O2(g)
HOMEWORK

Textbook page 364 - 366 ques. 16, 18, 20
VOLUME- VOLUME
12.3
LIMITING REAGENT AND
PERCENT YIELD
LIMITING REAGENT AND EXCESS REAGENT

Limiting Reagent is the reagent that
determines the amount of product that can be
formed by a reaction
H2 + O2  H2O
O2 – Red
H2 - Blue
LIMITING REAGENT AND EXCESS REAGENT

Excess Reagent- is the reagent not used up
(what is left over in a chemical reaction)
EXAMPLE
CHEMICAL EQUATION FOR THE PREPARATION OF
AMMONIA
EXAMPLE #1
PRACTICE ON YOUR OWN

Page 370 # 25
PERCENT YIELD

The percent yield is a measure of the efficiency of
a reaction carried out in the laboratory.
Theoretical yield -maximum amount of product that
could be formed from given amounts of reactants
 Actual yield the amount of product that actually forms
when the reaction is carried out in the laboratory

EXAMPLE #1 (PERCENT YIELD )
EXAMPLE #2
PERCENT YIELD
PRACTICE ON YOUR OWN
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Page 374-375
# 29 and 31