Section 9.1
Using Chemical Equations (Stoichiometry)
Unit 2: Chemical reactions
Mole Ratios & Stochiometry
Section 9.1
Using Chemical Equations (Stoichiometry)
Objectives
1. To understand the information given in a
balanced equation
2. To use a balanced equation to determine
relationships between moles of reactant and
products (mole ratio) (Stoichiometry)
Section 9.1
Using Chemical Equations (Stoichiometry)
Quinoa Side Dish
Section 9.1
Using Chemical Equations (Stoichiometry)
1. Information Given by Chemical Equations
• The coefficients of a balanced equation give
the relative numbers of molecules.
• A balanced chemical equation gives relative
numbers (or moles) of reactant and product
molecules that participate in a chemical
reaction.
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• A balanced equation can predict the moles of
product that a given number of moles of reactants
will yield. How many moles 02 from __ moles H2O?
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• The mole ratio allows us to convert from
moles of one substance in a balanced
equation to moles of a second substance in
the equation.
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
•
•
•
•
•
The mole ratio is determined by:
Mole ratio = new mole substance A
balanced mole substance A
How many mole ___ from ___ moles ___?
2H2O  2H2 + O2
• mole
mole ratio
mole
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• Must first have a BALANCED equation
• __C3H8(g) + __O2  __CO2(g) + __H2O(g) + heat
• Calculate the number of moles of CO2 formed
when 4.30 mol C3H8 combust.
• mole
mole ratio
mole
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• From this new mole value, we can calculate the
new recipe!
•
C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• 4.30 C3H8 +__O212.9 CO2 + __ H2O + heat
• Calculate the number of moles of H2O formed
when 4.30 mol C3H8 combust.
• mole
mole ratio
mole
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• For all reactants AND products!
•
C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• 4.30 C3H8 +__O212.9 CO2 + 17.2 H2O + heat
• Calculate the number of moles of O2 consumed
when 4.30 mol C3H8 combust.
• mole
mole ratio
mole
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• Calculate the number of moles of all reactants
and products when 2.70 mol C3H8 combust.
• 2.7 C3H8 +__O2 __ CO2 + __ H2O + heat
• 2.7 C3H8 +13.5 O2 8.1 CO2 + 10.8 H2O + heat
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• Balance the following equation and determine
how much NH3 can be made from 1.3 mol H2.
• __N2 + __H2  __NH3
•
N2 + 3H2  2NH3
• How much N2 will be needed to completely
react the 1.3 mol H2?
Section 9.1
Using Chemical Equations (Stoichiometry)
2. Mole-mole Relationships
• Using the balanced equation, determine the
new mole recipe given the new mole value.
•
2 K + 2 H2O 
H2 + 2 KOH
• ___K + 1.7 H2O  ___ H2 + ___KOH
• 1.7 K + 1.7 H2O  0.85 H2 + 1.7 KOH
•
4 NH3 + 5 O2  4 NO + 6 H2O
• ___ NH3 + ___ O2  0.6 NO + ___ H2O
• 0.6 NH3 + 0.75 O2  0.6 NO + 0.9 H2O
Section 9.1
Using Chemical Equations (Stoichiometry)
Objectives Review
1. To understand the information given in a
balanced equation
2. To use a balanced equation to determine
relationships between moles of reactant and
products (mole ratio) (Stoichiometry)
Section 9.2
Using Chemical Equations to Calculate Mass
Objectives
1. To learn to use Stoichiometry to relate
masses of reactants and products in a
chemical reaction
2. To perform mass calculations that involve
scientific notation
3. Calculate the theoretical predications for a
reaction that will be performed in the lab
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
• Stoichiometry is the process of using a
balanced chemical equation to determine the
relative masses of reactants and products
involved in a reaction.
– Scientific notation can be used for the
masses of any substance in a chemical
equation.
• Mole ratio = new mole substance A
•
balanced mole substance A
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
•
N2 + 3H2  2NH3
• How much NH3 will be produced when 2.6 g H2
completely reacts with N2?
• **Can’t do a mass ratio!!**
• Must first convert from grams to moles.
• Don’t forget the mole!
• grams
• mole
grams
mole ratio
mole
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
•
N2 + 3H2  2NH3
• How much NH3 will be produced when 2.6 g H2
completely reacts with N2?
• 2.6 grams H2
• mole
grams
mole ratio
• 0.867 mol NH3: 14.71 g NH3
mole
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
•
N2 +
3H2 
2NH3
•
__ N2 + 1.3 H2  0.867 NH3 (new mol recipe)
• __ g N2 2.6 g H2
14.71 g NH3 (new g recipe)
• How much N2 will be consumed when 2.6 g H2
completely reacts?
• Use the same mole ratio—
• 0.433 mol N2: 12.12 g N2
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
•
N2 +
3H2 
2NH3
• 0.433 N2 + 1.3 H2  0.867 NH3 (new mol recipe)
• 12.12 g N2 2.6 g H2
14.71 g NH3 (new g recipe)
• How does this relate to the Law of Conservation of
Mass?
• Does the mass reactants = mass products?
• 12.12 g + 2.6 g = 14.72 g
• Compared to 14.71 g Why the 0.01 g difference?
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
Section 9.2
Using Chemical Equations to Calculate Mass
1. Mass Calculations
•
__Al(S) + __I2(g)  __AlI3(s)
• Balance the equation and calculate the mass of
I2 needed to completely react with 35.0 g Al.
Also calculate the mass of AlI3 that will be
formed.
• grams
grams
• mole
mole ratio
mole (next)
Section 9.2
Using Chemical Equations to Calculate Mass
2. Scientific Notation
•
•
•
•
•
•
__LiOH(s) + __CO2(g)  __Li2CO3(s) + __H2O(l)
What mass of CO2 can 1 X 103 g LiOH absorb?
What mass of H2O will be available for use?
What mass of Li2CO3 will be generated?
First Balance! Then, below!!
grams
grams
• mole
mole ratio
mole
Section 9.2
Using Chemical Equations to Calculate Mass
2. Scientific Notation (CO2 scrubber on space vehicles)
•
•
•
•
•
•
•
•
•
2 LiOH(s) + CO2(g)  Li2CO3(s) +
H2O(l)
41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O
1 X 103 g
915.2 g
1539.2 g
374.4 g
What mass of CO2 can 1 X 103 g LiOH absorb?
What mass of H2O will be available for use?
What are some possible uses for the H2O?
What mass of Li2CO3 will be generated?
What will be done with the Li2CO3?
Conserved?
Section 9.2
Using Chemical Equations to Calculate Mass
2. Mass Calculations Using Scientific Notation
• Hydrofluoric acid, an aqueous solution
containing dissolved hydrogen fluoride, is used
to etch glass by reacting with silica, SiO2, in the
glass to produce gaseous silicon tetrafluoride
and liquid water. The unbalanced reaction is:
• __HF(aq) + __SiO2(s)  __SiF4(g) + __H2O(l)
• Calculate the mass of HF that is needed and
the products that are produced to react
completely with 5.68 g SiO2.
Section 9.2
Using Chemical Equations to Calculate Mass
2. Mass Calculations Comparisons
• Which is the better antacid? Baking Soda,NaHCO3,
Mg(OH)2?
• How many moles of HCl will react with 1.00 g of each
antacid?
• NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g)
Mg(OH)2(s) + 2HCl(aq)  2H2O(l) + MgCl2(aq)
• grams
• mole
grams
mole ratio
mole
Section 9.2
Using Chemical Equations to Calculate Mass
2. Mass Calculations for Lab- don’t round too much!
• Predict the charged balanced products and
their solubility. Then balance the equation, and
calculate the number of grams of all reactants
and products needed to completely react with
0.01 mol SrCl2.
• ___SrCl2 + ___Na2CO3
•
Section 9.2
Using Chemical Equations to Calculate Mass
2. Mass Calculations for lab
•
SrCl2 +
Na2CO3
SrCO3 + 2NaCl
• 0.01 SrCl2 + ___Na2CO3___SrCO3+___NaCl
• 0.01SrCl2 + 0.01Na2CO3  0.01SrCO3 + 0.02NaCl
•
1.59 g +
1.06 g
• Conservation of mass?
• Theoretical vs Real…

1.48 g
+
1.17 g
Section 9.2
Using Chemical Equations to Calculate Mass
2. Mass Calculations for lab
• Calculate the amount of grams contained in
0.01 mol the following hydrated crystals:
• SrCl2 x 6H2O
• Na2CO3 X H2O
3
0
Section 9.2
Using Chemical Equations to Calculate Mass
2. Mass Calculations for lab
• 0.01SrCl2 x 6H2O + 0.01Na2CO3 X H2O  0.01SrCO3 + 0.02NaCl
•
2.67 g
+
1.24 g
 1.48 g + 1.17 g
Section 9.2
Using Chemical Equations to Calculate Mass
Objectives Review
1. To understand the information given in a
balanced equation
2. To use a balanced equation to determine
relationships between moles of reactant and
products (mole ratio) (Stoichiometry)
3. Calculate the theoretical predications for a
reaction that will be performed in the lab
Section 9.3
Limiting Reactants and Percent Yield
Objectives
1. To understand the concept of limiting
reactants
2. To learn to recognize the limiting reactant in
a reaction
3. To learn to use the limiting reactant to do
stoichiometric calculations
4. To learn to calculate percent yield
Section 9.3
Limiting Reactants and Percent Yield
1. The Concept of Limiting Reactants
• Stoichiometric mixture (balanced)
– N2(g) + 3H2(g)  2NH3(g)
Section 9.3
Limiting Reactants and Percent Yield
1. The Concept of Limiting Reactants
• Limiting reactant mixture (runs out first)
– N2(g) + 3H2(g)  2NH3(g)
Section 9.3
Limiting Reactants and Percent Yield
Mixture of CH4 and H2O Molecules Reacting
CH4 + H2O  3H2 + CO
Section 9.3
Limiting Reactants and Percent Yield
CH4 + H2O  3H2 + CO
Section 9.3
Limiting Reactants and Percent Yield
Limiting Reactants
•
•
•
•
The amount of products that can form is
limited by the methane.
Methane is the limiting reactant.
Water is in excess.
Limiting and Excess Reactants depend
on the actual amounts of reactants
present and must be calculated for each
different reaction. (Water won’t always
be in excess!)
Section 9.3
Limiting Reactants and Percent Yield
Limiting Reactants
Section 9.3
Limiting Reactants and Percent Yield
2. Calculations Involving a Limiting Reactant
• What mass of water is needed to react with
249 g methane (CH4)?
• CH4 + H2O  3H2 + CO
• 249 g CH4
• How many g of H2 and CO are produced?
• 279 g H2O, 93 g H2, 434 g CO
Section 9.3
Limiting Reactants and Percent Yield
2. Calculations Involving a Limiting Reactant
• How many g of H2 and CO are produced
when 249 g methane (CH4) reacts with
300 g H2O?
• CH4 + H2O  3H2 + CO
• 249 g + 279 g  93 g + 434 g (last slide)
• 249 g + 300 g  93 g + 434 g
• The 249 g CH4 will react with only 279 g
H2O, thereby leaving 21 g H2O in excess.
CH4 is the reactant that will run out first and
is the LIMITING REACTANT.
Section 9.3
Limiting Reactants and Percent Yield
2. Calculations Involving a Limiting Reactant
• So, how do we figure this out using Stoich?
• 1) Balance Reaction
• 2) Do Stoich to relate EACH reactant to one
product
• 3) Whichever reactant produces the LEAST
amount of product is the LIMITING
REACTANT
Section 9.3
Limiting Reactants and Percent Yield
2. Calculations Involving a Limiting Reactant
Section 9.3
Limiting Reactants and Percent Yield
2. Calculations Involving a Limiting Reactant
• Suppose 2.50 X 104 g N2 reacts with 5 X 103 g H2.
Determine the limiting reactant.
• ___N2 + ___H2  ___NH3
• 2.50 X 104 g N2
• 5 X 103 g H2
• From N2  1785.72 mol NH3
• From H2  1666.67 mol NH3 ; 28333.3 g NH3
• Not just because original mass of H2 was less.
Section 9.3
Limiting Reactants and Percent Yield
2. Calculations Involving a Limiting Reactant
• How much N2 will be produced when 18.1 g NH3
and 90.4 g CuO react? Determine the limiting
reactant.
• __NH3 + __CuO  __N2 + __Cu + __H2O
• 18.1 g NH3
• 90.4 g CuO
• NH3 0.53 mol N2
• CuO 0.38 mol N2 *limiting reactant* 10.64 g N2
Section 9.3
Limiting Reactants and Percent Yield
3. Percent Yield
• Theoretical Yield
– The maximum amount of a given product that can be
formed when the limiting reactant is completely
consumed. Just what we’ve been calculating!
• The actual yield (amount produced is usually given) of a
reaction is usually less than the max theoretical yield
because of side or incomplete reactions.
• Percent Yield The actual amount of a given product as the
percentage of the theoretical yield.
Section 9.3
Limiting Reactants and Percent Yield
3. Percent Yield
• From the previous set of calculations, we
determined that 10.64 g of N2 would theoretically
be produced from the reaction:
• 2NH3 + 3CuO  N2 + 3Cu + 3H2O
• What would the Percent Yield be if 7.25 g of
N2 were actually produced?
•
•
7.25 g N2
10.64 g N2
X 100 = 68.14 %
Section 9.3
Limiting Reactants and Percent Yield
3. Percent Yield
• Consider the following reaction:
• __CO(g) +
__H2(g)  __CH3OH(l)
• If 6.85 X 104 g CO reacts with 8.6 X 103 g H2 to
produce an actual yield of 3.57 X 104 g CH3OH,
determine the limiting reactant, the theoretical yield,
and the percent yield.
Section 9.3
Limiting Reactants and Percent Yield
3. Percent Yield
• CO(g)
+
• 2446.4 mol CO
• 2446.4 mol CH3OH
2H2(g) 
CH3OH(l)
4300 mol H2
2150 mol CH3OH *Limiting
• 2150 mol CH3OH *32g/mol = 68800 g CH3OH
• % yield = 3.57 X 104 g/ 6.88 X 104 g = 51.89 %
• 28 g/mol CO: 2 g/mol H2: 32 g/mol CH3OH
Section 9.3
Limiting Reactants and Percent Yield
3. Percent Yield
• Consider the following reaction:
• __TiCl4
+
__O2  __TiO2 + __Cl2
• If 6.71 X 103 g TiCl4 reacts with 2.45 X 103 g O2
with a percent yield of 75%, determine the limiting
reactant, the theoretical yield, and the actual yield
of TiO2.
• TiCl4 is limiting
• 2119.6 g TiO2 is actually produced in this reaction.
• 189.68 g/mol TiCl4: 32 g/mol o2: 79.88 g/mol TiO2
Section 9.3
Limiting Reactants and Percent Yield
Objectives Review
1. To understand the concept of limiting
reactants
2. To learn to recognize the limiting reactant in
a reaction
3. To learn to use the limiting reactant to do
stoichiometric calculations
4. To learn to calculate percent yield
5. Work Session: Page 308 2, 3, 4, 5