Chemical Equations and Reaction
Stoichiometry
Chapter 3
Chemical Reactions
Reactant
Reactant
Product
Product
1 molecule CH4
2 molecules O2
1 molecule CO2
2 molecules H2O
1 mole CH4
2 moles O2
1 mole CO2
2 moles H2O
16.0 grams CH4
64.0 grams O2 (2
 32.0 g)
44.0 g CO2
36.0 grams H2O
(2  18.0 g)
Physical State Representation
• (g) = _________
• (l) = _________
• (s) = _________
• (aq) = _______________
These representations will commonly be
observed in chemical equations.
Chemical Equations
• There is no loss in the quantity of matter from a
chemical reaction. This is the _____ __
__________ __ _____.
• What is the total mass on both sides of the
chemical reaction shown previously?
• A balanced chemical equation must have the
_____ of each kind of atom on both sides of the
equation.
– Smallest possible whole-number coefficients.
Balancing Chemical Equations
•
•
•
•
__Al(s) + __Br2(l)  ___Al2Br6(s) (covalent)
__C3H8(g) + __O2(g)  __CO2(g) + __H2O(g)
__C4H10(l) + __O2(g)  __CO2(g) + __H2O(g)
__P4(s) + __Cl2(g)  __PCl3(s)
– Demo: On demo. CD
Be sure to balance equations before proceeding with
calculations.
Calculations Based on Balanced
Chemical Equations
• The ______ in a balanced equation
represent the relative ratio of moles or
molecules (formula units).
• Calculations utilize a series of _________
to obtain the desired answer.
– Make sure that the ________ cancel.
Calculations Based on Balanced
Chemical Equations
• Conversion process
– Convert grams to moles (using molar mass)
– Convert moles to moles (from balanced
equation)
• This can also be molecules. Why?
– Convert moles to grams (using molar mass)
This is the general steps that will be followed to
solve equations.
Solving Problems from Chemical
Equations
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
• How many CH4 molecules required to react
completely with 82 molecules of O2?
__C4H10(l) + __O2(g)  __CO2(g) + __H2O(g)
• How many O2 molecules required to react
completely with 48 molecules of C4H10?
Solving Problems from Chemical
Equations
•
•
•
•
__Fe2O3(s) + __CO(g)  __Fe(s) + __CO2(g)
How many moles of Fe2O3 are required to react
completely with 4.9 moles of CO?
What mass of CO2 can be produced from 0.540
moles of Fe2O3 with excess CO?
__P4(s) + __Cl2(g)  __PCl3(s)
How many moles of PCl3 can be produced from
1.48 moles of Cl2(g)? Assume the P4 is in excess.
How many grams of PCl3 can be produced from
42.3 grams of P4?
More Problems??
__NH3 + __O2  __NO + __H2O
• How many grams of NO can be produced
from 17.80 grams of O2? NH3 is in excess.
• How many molecules of NH3 are required
to produce 7.31  10-10 grams of H2O?
Limiting Reactant
• In a typical reaction, there is one substance that is
generally used up first. This reactant is termed as
the ‘limiting reactant’ since is limits how much
product will be formed. The reactants will be in
excess.
DEMO: Setting a table for six people using 10
plates, 8 spoon, and 5 forks.
What is limiting and what is in excess? This is the
same with a reaction.
Limiting Reactant
DEMO: CD (5-4) NH3 formation
• Method for calculating the limiting reactant
– Calculate the amount of product that would be
formed from each reactant.
– The one that produces the least amount of
product is the limiting reactant/reagent.
Limiting Reactant
• What is the maximum amount of sulfur dioxide that
can be produced from 95.6 grams of carbon
disulfide and 110 grams of oxygen?
CS2 + O2  CO2 + SO2
• What is the maximum amount of HCl that can be
produced from 28.2 grams of NaCl, 23.8 grams of
H2O, and 32.1 grams of SiO2?
NaCl + H2O + SiO2  HCl + Na2SiO3
Limiting Reactant
• Determine the amount of Cu3P formed if
225 grams of copper is reacted with 65.0
grams of P4. What is the limiting reactant?
– ___Cu + ___P4 ____Cu3P
Determining Percent Yields
• In all the previous calculations, it is
assumed that the yield is 100%. In the real
world, however, the actual yield is always
below 100%. Why?
– Many reactions do not go to completion.
– Two or more reaction pathways occur
simultaneously.
– Some of the product is lost upon isolation.
Determining Percent Yields
• Percent yield
actual yield
percenty yield 
 100 %
theoretica l yield
• A 10.0-gram sample of ethanol, C2H5OH, was
reacted with excess CH3COOH to produce 14.8
grams of CH3COOC2H5. What is the percent
yield?
CH3COOH + C2H5OH  CH3COOC2H5 + H2O
Determining Percent Yields
• Aspirin can be synthesized as shown below.
For the reaction 12.4 grams of C4H6O3 is
reacted with an excess of C7H6O3. The
amount of aspirin, C9H8O4, acquired from the
reaction is 26.28 grams. What is the percent
yield?
2C7H6O3(s) + C4H6O3(l)  2C9H8O4(s) +
H2O(l)
Concentrations of Solutions
• In many chemical reactions, the reactants are that
mixed are present as solutions.
• Solution consists of a substance (solute) dissolved
in another substance (solvent).
– DEMO: CuSO4(aq) and Ca(NO3)2(aq)
– Mix Ca(NO3)2 and Na2(CO3)
• Ca(CO3) is the solid. Write the chemical equation.
• Solutions are _________ mixtures. Mixing occurs
at the molecular level.
– Aqueous solutions – solvent in ______.
Concentration of Solutions
• Concentration expresses the amount of solute
dissolved in a give amount of solvent or solution.
• Percent by mass of solute
percent solute 
mass of solute
 100 %
mass of solution
mass of solution  mass of solute  mass of solvent
• What mass of NaCl is required to prepare 75.0
grams of a 5.00% solution of NaCl?
DEMO: Prepare this solution.
Concentrations of Solutions
• A 7.50% solution of C6H12O6 contains 50.0
grams of C6H12O6. What is the mass of the
solution?
• Calculate the mass of NaOH in 150.0 mL of
an 8.00% solution of NaOH. The density of
the solution is 1.09 g/mL.
Concentrations of Solutions
• Molarity is defined as the number of moles of
solute per liter of solution.
number of moles of solute
molarity 
number of liters of solution
DEMO: 1.0 and 0.1 M solution of CuSO4 How
many grams of CuSO4 required to make 1 liter of
these solutions?
• Prepare 50.0 mL of a 0.100 M solution of aqueous
K2Cr2O7.
• What is the molarity of a solution that contains
4.91 grams of HNO3 in 500.0 milliliters of
solution?
Concentration of Solutions
• Determine the mass of BaSO4 required to
produce 250.0 milliliters of a 0.125 M
solution of barium sulfate.
• Optional: The density of a concentrated
HCl solution is 1.185 g/mL and it is 36.31%
by mass. What is the molarity?
Solution by Dilution
• When a specified amount of concentrated
solution is diluted, the molarity decreases and
the volume increases. The number of moles,
however, remains constant.
– Concentrated solution
• Number of moles = M1  V1
– Diluted solution
• Same number of moles = M2  V2
Therefore, M1  V1 = M2  V2
Solutions by Dilution
• M1  V1 = M2  V2 This relationship is used to
calculate the amount of concentrated solution, V1,
needed to prepare a diluted solution of given volume,
V2, and molarity, M2.
• How would you prepare 25.0 mL of a 8.00  10-3 M
solution of K2Cr2O7 from 0.100 M K2Cr2O7?
• How many mL of 8.0 M sulfuric acid is required to
produce 50.0 mL of a 0.150 M solution of sulfuric
acid?
Solutions in Chemical Reactions
• In many chemical reaction it is beneficial or
even necessary to add reactants in solution.
If molarity is known, the amount of solute
(or reactant) to be added to a reaction can be
determined.
• What volume of 0.500 M BaCl2 is required
to react completely with 4.32 grams of
Na2SO4?
Solutions in Chemical Reactions
• What volume of 0.200 M NaOH will react
with 50.0 mL of 0.200 M aluminum nitrate?
What will be the mass of the aluminum
hydroxide precipitate?