Chemical Reactions
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Na(s) + H2O(l ) 
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Na(s) + H2O(l )  NaOH(aq) + H2(g)
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Na(s) + H2O(l )  NaOH(aq) + H2(g)

The equation is not yet balanced. Hydrogens
come in twos on the left, and three hydrogens
are on the right side of the equation.
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Na(s) + H2O(l )  NaOH(aq) + H2(g)

Try a “2” in front of the water.
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Na(s) + 2 H2O(l )  NaOH(aq) + H2(g)

We now have two O atoms on the left, so we
need to put a 2 before NaOH.
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)

The two sodium atoms on the right require that
we put a 2 in front of Na on the left.
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)

The two sodium atoms on the right require that
we put a 2 in front of Na on the left. The
equation is now balanced.
Balancing Chemical EquationsProblem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen.
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
Left Side
Right Side
Na- 2
Na- 2
H- 4
H- 4
O- 2
O- 2
Chemical Equations
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
The balanced chemical equation can be
interpreted in a variety of ways.
It could say that 2 atoms of sodium react
with 2 molecules of water to produce 2
molecules of sodium hydroxide and a molecule
of hydrogen.
Chemical Equations
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
The balanced chemical equation can be
interpreted in a variety of ways.
It could say that 200 atoms of sodium react
with 200 molecules of water to produce 200
molecules of sodium hydroxide and 100
molecules of hydrogen.
Chemical Equations
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
The balanced chemical equation can be
interpreted in a variety of ways.
 It is usually interpreted as 2 moles of sodium
will react with 2 moles of water to produce 2
moles of sodium hydroxide and 1 mole of
hydrogen.
 The balanced equation tells us nothing about the
masses of reactants or products.
Types of Chemical Reactions
A combination reaction is a reaction with two or
more reactants, and a single product.
2 CO (g) + O2(g)  2 CO2(g)
Types of Chemical Reactions
A decomposition reaction is a reaction has one
reactant, and two or more products.
CaCO3(s)  CaO(s) + CO2(g)
Stoichiometry


Stoichiometry is a Greek word that means using
chemical reactions to calculate the amount of
reactants needed and the amount of products
formed.
Amounts are typically calculated in grams (or
kg), but there are other ways to specify the
quantities of matter involved in a reaction.
Stoichiometry
A balanced chemical equation or reaction is
needed before any calculations can be made.
The formulas of all reactants and products
are written before attempting to balance the
equation.
Stoichiometry Problem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen. How
many grams of sodium are needed to produce
50.0g of hydrogen?
A balanced chemical equation is needed
before any calculations can be made.
Stoichiometry Problem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen. How
many grams of sodium are needed to produce
50.0g of hydrogen?
Na(s) + H2O(l )  NaOH(aq) + H2(g)
Stoichiometry Problem

Sodium metal reacts with water to produce
aqueous sodium hydroxide and hydrogen. How
many grams of sodium are needed to produce
50.0g of hydrogen?
2 Na(s) +2 H2O(l )  2 NaOH(aq) + H2(g)
Stoichiometry Problem

2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
How many grams of sodium are needed to
produce 50.0g of hydrogen?
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
Stoichiometry Problem


2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
Although the question doesn’t state it, you can
assume enough water is present for complete
reaction.
We can map out the problem:
g H2  moles H2 moles Na  grams Na
Stoichiometry Problem

2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
We can map out the problem:
g H2  moles H2 moles Na  grams Na
We use the molar mass of H2 to go from grams
of H2 to moles of H2.
Stoichiometry Problem

2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
We can map out the problem:
g H2  moles H2 moles Na  grams Na
molar mass of H2
Stoichiometry Problem


2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
We can map out the problem:
g H2  moles H2 moles Na  grams Na
molar mass of H2
We use the coefficients from the balanced equation to
go from moles of H2 to moles of Na.
Stoichiometry Problem

2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
We can map out the problem:
g H2  moles H2 moles Na  grams Na
molar mass of H2 coefficients
Stoichiometry Problem


2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
We can map out the problem:
g H2  moles H2 moles Na  grams Na
molar mass of H2 coefficients
We use the molar mass of Na to go from moles
of Na to grams of Na.
Stoichiometry Problem

2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
We can map out the problem:
g H2  moles H2 moles Na  grams Na
molar mass of H2 coefficients molar mass of Na
Stoichiometry Problem
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
? grams
50.0 g
g H2  moles H2 moles Na  grams Na
molar mass of H2 coefficients molar mass of Na
(50.0 g H2) (1 mol H2) (2 moles Na) ( 22.99 g Na)
(2.02 g H2) (1 mol H2) (1 mol Na)
Stoichiometry Problem
2 Na(s) + 2 H2O(l ) 2NaOH(aq) + H2(g)
(50.0 g H2) (1 mol H2) (2 moles Na) ( 22.99 g Na) =
(2.02 g H2) (1 mol H2) (1 mol Na)
= 1,138 grams Na = 1.14 x 103 g Na = 1.14 kg Na
Limiting Reagent Problems

Sometimes you are given quantities of more
than one reactant, and asked to calculate the
amount of product formed. The quantities of
reactants might be such that both react
completely, or one might react completely, and
the other(s) might be in excess. These are called
limiting reagent problems, since the quantity of
one of the reacts will limit the amount of
product that can be formed.
Limiting Reagent - Problem

Aluminum and bromine react to form aluminum
bromide. If 500. g of bromine are reacted with
50.0 g of aluminum, what is the theoretical yield
of aluminum bromide?
1. First write the formulas for reactants and products.
Al + Br2  AlBr3
Limiting Reagent - Problem

Aluminum and bromine react to form aluminum
bromide. If 500. g of bromine are reacted with
50.0 g of aluminum, what is the theoretical yield
of aluminum bromide?
2. Now balance the equation by adding coefficients.
2 Al +3 Br2 2 AlBr3
Limiting Reagent - Problem

Aluminum and bromine react to form aluminum
bromide. If 500. g of bromine are reacted with
50.0 g of aluminum, what is the theoretical yield
of aluminum bromide?
2 Al +3 Br2 2 AlBr3
The theoretical yield is the maximum amount of
product that can be formed, given the amount of
reactants. It is usually expressed in grams.
Limiting Reagent - Problem
Given:

2 Al +3 Br2 2 AlBr3
50.0g 500.g
? grams
There are several ways to solve this problem. One
method is to solve the problem twice. Once, assuming
that all of the aluminum reacts, the other assuming that
all of the bromine reacts. The correct answer is
whichever assumption provides the smallest amount of
product.
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:

50.0g
500.g
? grams
The problem can be mapped:
Al: grams Al  moles Al moles AlBr3  g AlBr3
molar mass Al
coefficients
molar mass AlBr3
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:

50.0g 500.g
? grams
The problem can be mapped:
Al: grams Al  moles Al moles AlBr3  g AlBr3
molar mass Al
coefficients
molar mass AlBr3
(50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr3) (266.7 g AlBr3)
(26.98 g Al) ( 2 mol Al)
(mol AlBr3)
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:
50.0g
500.g
? grams
Al: (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr3) (266.7 g AlBr3)
(26.98 g Al) ( 2 mol Al)
= 494. g AlBr3 (if all of the Al reacts)
(mol AlBr3)
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:
50.0g
500.g
? grams
The calculation is repeated for Br2.
g Br2 moles Br2  moles AlBr3  g AlBr3
molar mass Br2 coefficients
molar mass AlBr3
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:
50.0g
500.g
? grams
The calculation is repeated for Br2.
g Br2 moles Br2  moles AlBr3  g AlBr3
molar mass Br2 coefficients molar mass AlBr3
(500. g Br2) (1 mol Br2) (2 moles AlBr3) (266.7 g AlBr3)
(159.8 g Br2) (3 moles Br2) (1 mol AlBr3)
= 556. grams of AlBr3
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:

50.0g
500.g
? grams
Summary:
We have enough Al to produce 494. g AlBr3
We have enough Br2 to produce 556. grams of AlBr3
The theoretical yield is 494. grams of AlBr3.
Limiting Reagent - Problem



2 Al +3 Br2 2 AlBr3
Given: 50.0g 500.g
494. g
Summary:
All of the Al reacts, so Al is limiting.
Bromine is in excess.
Additional questions:
1. How much bromine is left over?
2. If 418 grams of AlBr3 is obtained, what is
the % yield?
Limiting Reagent - Problem
2 Al +3 Br2 2 AlBr3
Given:
50.0g 500.g
494. g
1. How much bromine is left over?
Since all 50.0 g of the Al reacts, the product
must contain 494.g -50.g = 444. g of
bromine. Therefore, 500.g- 444.g = 56. g of
Br2 are left over.
Limiting Reagent - Problem
Given:
2.
2 Al +3 Br2 2 AlBr3
50.0g 500.g
494. g
If 418 grams of AlBr3 is obtained, what is the %
yield?
The percent yield is (actual yield) (100%)
(theoretical yield)
% yield = (418 g) (100%) = 84.6 %
(494g)
Types of Chemical Reactions
In a combustion reaction, a substance burns in the
presence of oxygen. If the reactant is a
compound containing on carbon, hydrogen and
oxygen, the products are water and carbon
dioxide.
CH4(g) +2 O2(g)  2 H2O (l) + CO2(g)
Alkali Metal Reactions
The group IA metals will donate electrons to
non-metals. Reactions with the halogens (group
VIIA) produces ionic salts.
2 Na(s) + Cl2(g)  2 NaCl(s)
Reactions with water produces hydrogen gas
and the metal hydroxide.
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
Reactions of the Halogens
The elements of group VIIA readily pick up
electrons from metals to form ionic compounds
(salts).
In addition, the halogens react with
hydrogen gas to form covalently bonded hydrogen
halides.
H2(g) + Cl2(g)  2 HCl(g)
Reactions of the Halogens
The halogens can also react with each other
to form interhalogen compounds. The products are
covalently bonded.
Br2(g) + F2(g)  2 BrF(g)