Transcript Stoichiometry

Stoichiometry
SOLVING PROBLEMS BASED ON THE
BALANCED CHEMICAL EQUATION
2Al + 3FeO  Al2O3 + 3Fe
 In this balanced equation, four chemical substances
are represented.
 Two Reactants: aluminum and iron(II) oxide
 Two Products: aluminum oxide and iron
2Al + 3FeO  1Al2O3 + 3Fe
 Also the ratio (by number) in which they react and
are produced is indicated by the coefficients.
2Al + 3FeO  1Al2O3 + 3Fe
 However, a reactant or product coefficient is not a
part of the formula for the reactant or product.
2Al + 3FeO  1Al2O3 + 3Fe
 The formula for:
 aluminum is Al, not 2Al

(Molar Mass based on 1 Al atom)
2Al + 3FeO  1Al2O3 + 3Fe
 The formula for:
 aluminum is Al, not 2Al


(Molar Mass based on 1 Al atom)
iron(II) oxide is FeO, not 3FeO

(Molar mass based on 1 Fe atom and 1 O atom)
2Al + 3FeO  1Al2O3 + 3Fe
 The formula for:
 aluminum is Al, not 2Al


iron(II) oxide is FeO, not 3FeO


(Molar Mass based on 1 Al atom)
(Molar mass based on 1 Fe atom and 1 O atom)
Aluminum oxide is Al2O3, not 1Al2O3

(Molar mass based on 2Al atoms and 3 O atoms)
2Al + 3FeO  1Al2O3 + 3Fe
 The formula for:
 aluminum is Al, not 2Al


iron(II) oxide is FeO, not 3FeO


(Molar mass based on 1 Fe atom and 1 O atom)
Aluminum oxide is Al2O3, not 1Al2O3


(Molar Mass based on 1 Al atom)
(Molar mass based on 2Al atoms and 3 O atoms)
Iron is Fe, not 3Fe

(Molar mass is based on 1 Fe atom)
2Al + 3FeO  1Al2O3 + 3Fe
 Typically, stoichiometry problems involve questions
about amounts of one reactant or product based on
the reaction of, or the production of one of the other
substances.
 For example:

Based on the above equation, how much aluminum oxide
should be produced when a given amount of aluminum reacts.
2Al + 3FeO  1Al2O3 + 3Fe
 For these types of problems we use the coefficients to
help us make the determination.
 So for this particular problem, this:
 2Al + 3FeO  1Al2O3 + 3Fe
 Becomes:
 2Al + 3FeO  1Al2O3 + 3Fe
Based on the above equation, how much aluminum oxide should be
produced when a given amount of aluminum reacts
 2Al + 3FeO  1Al2O3 + 3Fe
 We focus on the 2 : 1 ratio for this problem.
 We know that the moles of Al reacted and moles of
Al2O3 produced must be in this 2:1 ratio.
Based on the above equation, how many mol of aluminum oxide should be
produced when 25.23 mol of aluminum react?
 2Al + 3FeO  1Al2O3 + 3Fe
 We focus on the 2 : 1 ratio for this problem.
 So x = 12.62 mol Al2O3
 (4 significant digits because the 2 and 1 are known exactly)
2Al + 3FeO  1Al2O3 + 3Fe
 Since the coefficient ratio is a ratio by number of
particles, and since moles is a measure of the count
of the number of particles, these coefficients will
always work when amounts are in moles.
2Al + 3FeO  1Al2O3 + 3Fe
 Gas volumes (when measured at the same conditions
of P and T) may also be used in conjunction with
coefficient ratios. This is because equal volumes of
different gases contain the same number of
molecules. (Thanks to Avogadro)
2Al + 3FeO  1Al2O3 + 3Fe
 So moles is a direct count and gas volume is directly
related to count, which is why both will work.
2Al + 3FeO  1Al2O3 + 3Fe
 However, we can’t measure the moles of aluminum
directly.
 In the lab, the most common method of measuring
the amounts of solid chemicals is by finding the
mass.
2Al + 3FeO  1Al2O3 + 3Fe
 The ratios by mass are not equal to the ratios by
number.

Example (2 elephants : 1 ant) is 2:1 by count, but not by mass.
 So if we are using mass, we’ll need to convert to
moles before using the coefficient ratio.
Based on the above equation, how many mol of aluminum oxide should be
produced when 25.23 grams of aluminum react?
 2Al + 3FeO  1Al2O3 + 3Fe
 First we change the mass to moles using molar mass.
Note that the molar mass of a substance is based on
its formula and is independent of its coefficient in
the balanced equation.
 Molar mass:




Al = 26.989 g/mol
FeO = (55.85 + 16.00) = 71.85 g/mol
Al2O3 = (2(26.98) + 3(16.00))= 101.96 g/mol
Fe = 55.85 g/mol
Based on the above equation, how many mol of aluminum oxide should be
produced when 25.23 grams of aluminum react?
 2Al + 3FeO  1Al2O3 + 3Fe
 First we change the mass to moles using molar mass.
Note that the molar mass of a substance is based on
its formula and is independent of the coefficient in
the balanced equation.

25.23 g/26.98g/mol = 0.9351 mol Al
 Now we can solve the problem using the
coefficients:
Based on the above equation, how many mol of aluminum oxide should be
produced when 25.23 grams of aluminum react?
 2Al + 3FeO  1Al2O3 + 3Fe


X= 0.4676 mol Al2O3
This answer can be changed to grams (if necessary) by using
the molar mass of Al2O3 . Remember that the coefficients are
only used for mole ratio when calculating molar mass.
2Al + 3FeO  1Al2O3 + 3Fe
Substance
A
Don’t use
coefficient
Mole
Ratio
Use
coefficients
Substance B
Don’t use
coefficient
2Al + 3FeO  1Al2O3 + 3Fe
 Mole Chart
 Shows amounts in Moles (or volumes for gases at same T & P)
I = amounts before reaction starts
 D = changes in amounts due to reaction (these changes depend on
how much of the starting amounts actually react) Must occur in
the ratio indicated by the coefficients.
 F = amounts remaining when reaction stops (usually when
limiting reactant has been consumed)

i
D
f
0.600
0.800
0
0
2Al + 3FeO  1Al2O3 + 3Fe
 Mole Chart
 Shows amounts in Moles (or volumes for gases at same T & P)
I = amounts before reaction starts
 D = changes in amounts due to reaction (these changes depend on
how much of the starting amounts actually react) Must occur in
the ratio indicated by the coefficients.
 F = amounts remaining when reaction stops (usually when
limiting reactant has been consumed)

i
D
f
0.600
0.800
-0.800
0
0
2Al + 3FeO  1Al2O3 + 3Fe
 Mole Chart
 Shows amounts in Moles (or volumes for gases at same T & P)
I = amounts before reaction starts
 D = changes in amounts due to reaction (these changes depend on
how much of the starting amounts actually react) Must occur in
the ratio indicated by the coefficients.
 F = amounts remaining when reaction stops (usually when
limiting reactant has been consumed)

i
0.600
0.800
0
0
D
0.533
-0.800
+0.267
+0.800
f
2Al + 3FeO  1Al2O3 + 3Fe
 Mole Chart
 Shows amounts in Moles (or volumes for gases at same T & P)
I = amounts before reaction starts
 D = changes in amounts due to reaction (these changes depend on
how much of the starting amounts actually react) Must occur in
the ratio indicated by the coefficients.
 F = amounts remaining when reaction stops (usually when
limiting reactant has been consumed)


i
0.600
0.800
0
0
D
-0.533
-0.800
+0.267
+0.800
f
0.067
0
0.267
0.800
Final amounts (f) = (i+D) for each column