Chapter 5
Chemical Reactions
Chemical equations describe CHEMICAL REACTIONS.
•During a chemical reaction, the ways in which atoms are joined together are
changed.
•OLD bonds are broken and NEW bonds are formed as REACTANTS are
converted into PRODUCTS.
•A reaction STARTS with substances known as REACTANTS (always written
on the LEFT side of the equation).
• The substances that are FORMED during a reaction are known as PRODUCTS
(always written on the RIGHT side of the equation).
• Chemical Equations are used (as chem. shorthand) to represent what is
occurring during a chemical reaction.
Ex.: Propane burns in oxygen to produce carbon dioxide gas and water vapor.
C3H8 + O2 --> CO2 + H2O
Why balance Equations?
• When 16.90g of the compound CaS is decomposed into its constituent elements,
the Ca and S produced has an identical mass of 16.90g.
Balancing Equations
•Make sure the chemical formulas are CORRECT (you cannot change the
correct formula of a substance).
•Use COEFFICIENTS (numbers in front of formulas) to change the number
of atoms of an element.
•Coefficients represent MULTIPLES of the formulas
Balance the following
N2 (g) +
H2 (g) --->
FeCl3 (s) --->
NH3 (g)
Fe(s) +
Cl2 (g)
C3H8 + O2 --> CO2 + H2O
C3H6 + O2 --> CO2 + H2O
Measuring MATTER
Counting
Weighing
Measuring Volume
- Avogadro’s #
- Molar Mass
- Molar Volume
Types of Chemical Particles
•Atoms - represented by the symbol of an element
•Molecules - represented by multiple symbols of NONMETAL elements
•Ions - represented by the formula of an ion
•Formula Units - represented by the formula of an IONIC compound.
•1 mole = 6.02 x 1023 particles
(a conversion factor)
6.02 x 1023 = Avogadro’s Number
•This number can be determined experimentally several ways:
–Measurement of crystal structure
•Ti - body-centered unit cell (2 atoms/unit cell)
•
# of atoms/mol = 6.02 x 1023atoms
MOLAR MASS
•The number of grams of a substance equivalent to the sum of all its average
atomic mass units (amu) is known as the molar mass.
•One mole of particles is equal to its molar mass in grams.
MOLAR VOLUME
•The volume, 22.4 L, of any gas at STP is known as the molar volume.
•STP = Standard Temperature and Pressure
273 K (O˚C, 32 ˚F) and 1 atm (101.3 kPa, 760 mm Hg, 29.92 in Hg)
Molar Conversions
particles
{
atoms
molecules
ions
formula u nits
6.023x10 23
moles
mole wt.
grams (mass)
22.4 lit ers of a gas (STP)
liters (volume)
STOICHIOMETRY
• Balanced chemical equations can be used to predict the QUANTITATIVE
amounts of REACTANTS and PRODUCTS.
N2(g) + 3H2 (g) ---> 2NH3 (g)
Moles
Particles
Molecules
Atoms
Mass
N2(g) + 3H2 (g) ---> 2NH3 (g)
How many moles of hydrogen will react with 0.00326 mol N2?
How many molecules of ammonia are produced when 4.55 x 1018 molecules of
hydrogen react?
How many atoms of hydrogen are produced?
What volume of ammonia can be manufactured when 6.5 liters of nitrogen react?
How many grams of nitrogen are required to react with 75.8 g hydrogen?
Stoichiometry
• Use the stoichiometric mole ratio to convert from moles of one substance to
moles of another substance within the reaction
C4H10 + O2 ---> CO2 + H2O
How many moles of water are produced when 0.48 moles of butane, C4H10, react?
How many molecules of butane are needed to produce 12.00 liters of CO2 at STP?
Stoichiometry
•In the lab, we determine the masses of different substances, rather than moles,
and therefore must be able to convert from grams of one substance in a reaction
to grams of another.
X + Y --> Z
•grams Y --> moles Y --> moles Z --> grams Z
NaOCl + KI + HC2H3O2 ---> I2 + NaCl + KC2H3O2 + H2O
•How many grams of iodine are produced when 0.35 g of potassium iodide react?
•grams KI --> mol KI --> mol I2 --> grams I2
I2 (aq) + Na2S2O3(aq) --> Na2S4O6(aq) + NaI
•How many grams of sodium iodide are produced when 0.203 g of iodine react?
•Which reactant limits the amount of product that can be made? Why?
Limiting Reactants
• The product(s) of a reaction is/are limited by how much of each reactant is
present (available) in the reaction.
•Two types of reactants
Limiting - this is the reactant you run out of first!
Excess - at the end of the reaction there will be some of this reactant left
over (excess:-)).
Combustion of Magnesium
Mg(s) + O2(g in air) --> MgO(s)
•How
much magnesium oxide can be produced when 1.085 g Mg burns in air?
•Determine the limiting and excess reactants
•Calculate
the theoretical yield from the limiting reactant
Mg(s) + O2(g in air) --> MgO(s)
Reaction of barium chloride & ammonium thiocyanide
BaCl2 + NH4SCN --> Ba(SCN) 2 + NH4Cl
Molar Mass
208.3g/mo l
76.1 g/mol
•34.5 g BaCl2 react with 44.3 g NH4SCN.
•How much NH4Cl can be produced (theoretical yield)?
•STEP 1: Determine the LR and ER! (use mole ratio)
•STEP 2: Determine the Theoretical Yield from LR
Experimental Reaction Yield
•Balanced equations can be used to calculate the amount of product that will
form during a reaction - called the THEORETICAL YIELD
•The amount of product that actually forms during a chemical reaction is
called the ACTUAL YIELD
•The actual yield is often less than the theoretical yield.
Percent Yield
•The percent yield is the ratio of the actual yield compared to the theoretical
yield, converted to a percent.
actual yield
•% YIELD = --------------------- x 100
theoretical yield
N2(g) + H2(g) ---> NH3 (g)
•0.075 g N2 react with 0.0095 g H2 to produce 0.051 g NH3.
•Which reactant is the limiting reactant? Which reactant is in excess?
•How much ammonia should be produced?
•What is the percent yield for this reaction?