Download CHEMICAL REACTIONS AND CHEMICAL EQUATIONS

CHEMICAL REACTIONS AND CHEMICAL EQUATIONS
Chemical reaction – is a process in which a substance is changed into one or more new
substances.
Chemical Equation – is a written statement that uses chemical symbols and chemical formulas
instead of words to describe the changes that occur in a chemical reaction.
A chemist’s shorthand description of a reaction.
2H2 + O2
Reactants
2H2O
Products
Reactant – is a starting material in a chemical reaction that undergoes change in the chemical
reaction.
Product – is a substance produced as a result of the chemical reaction
Conventions Used in Writing Chemical Equations
1.
2.
3.
4.
The correct formulas of the reactants are always written on the left side of the equation.
The correct formulas of the products are always written on the right side of the
equation.
The reactants and products are separated by an arrow pointing toward the products
Plus signs are used to separate different reactants or different products.
Chemical Stoichiometry – is the quantitative study of the relationships among the reactants and
products in a chemical reaction.
the stoichiometry of a chemical reaction always involves the molar relationships
between reactants and products and thus is given by the coefficients in a balanced
equation for the chemical reaction.
Mole method – the stoichiometric coefficients in a chemical equation can be interpreted as the
number of moles of each substance.
General approach for solving stoichiometry problems:
1.
2.
3.
4.
Write a balanced equation for the reaction.
Convert the given amount of the reactant (in grams or other units) to number of
moles.
Use the mole ratio from the balanced equation to calculate the number of moles of
product formed.
Convert the moles of products to grams of products.
Example.
2CO(g) + O2(g)
2CO2(g)
If 4.8 moles of CO react completely with O2 to form CO2, calculate the amount of CO2 produced
in moles.
= 4.8 mol CO2
Suppose 10.7 grams of CO react completely with CO2 to form CO2. How many grams of CO2 will
be formed?
Grams of CO
moles of CO
moles of CO2
grams of CO2
STOICHIOMETRY
INTERPRETING A CHEMICAL EQUATION
Quantitative Interpretation of Chemical Reactions
Stoichiometry is one of the most important topics in chemistry. It involves the use of the
chemical formulas , mole calculations, and chemical equations. Stoichiometry is also essential in
industry , there, it is used to do cost and analysis for manufacturing chemicals. In fact,
manufacturing processes are financed according to the cost of reactants and the values of
products are determined by stoichiometric calculations.
STOICHIOMETRY : The relationship of quantities ( mass of substance or volume of gas ) in a chemical
change according to the balanced chemical equation.
LAW OF COMBINING VOLUMES
The principle that volume of gases that combine in a chemical reaction, at the same
temperature and pressure, are in the ratio of small whole numbers, also called Gay-Lussac’s Law of
combining volumes.
In one of his experiment, Gay-Lussac studied the reaction of hydrogen and chlorine and found
that one volume of gaseous hydrogen combined with of gaseous chlorine to give two volumes of
hydrogen chloride gas. That is,
Hydrogen gas
+
chlorine
1 volume
Ex. 50.0 ml
→
hydrogen chloride gas
1 volume
2 volumes
50.0 ml
100.0 ml
In 1811, the Italian physicist Amadeo Avogadro explained the law of volumes. Avogadro
proposed that equal volume s of gas, at the same temperature and pressure, contain the same
number of molecules. That is any two gases containing the same numbers of molecules, will
occupy equal volumes. The statement is known as Avogadro’s Law. From the law, it was clear that
gases react in small whole number ratios because molecules of two gases react in small whole
number ratios.
Let’s consider the reaction of hydrogen and chlorine gases at the molecular level we have,
H2
+
Cl2
→ 2HCl
1 molecule
1 molecule
2 molecule
Avogadro’s law claims that equal number of H2 molecules and Cl2 molecule would have equal
volumes.
In another experiment Gay-Lussac study the reaction of hydrogen and oxygen. He found that
two volumes of hydrogen gas combined with one volume of oxygen gas. In this case, the gases
combine in the small whole number ratio of 2:1 that is,
Hydrogen
+
oxygen
→
water
2 volumes
1 volume
2 volume
100.0 ml
50.0 ml
100.0 ml
We can also look at the reaction of hydrogen and oxygen molecules
2 H2
+
O2
→
2 H2
Applying Avogadro’s law, there are twice the numbers of H 2 as O2 molecules , therefore, the
volume of H2 would be twice the volume of O2. Since the coefficients of H2O and H2 are the same
they would have equal volumes. Thus, Avogadro’s law neatly explained Gay-Lussac’s Law of
combining volumes. That is, volume of gases combines in small, whole –number ratios because
equal volumes of gases contain the same number of molecules.
Example Exercises:
1. After balancing the following equation for the combination of propane and oxygen, interpret
the coefficients in terms of a) moles and b) liters
C3H8
+
O2
→ CO2
+
H2O
Solution :
We can supply the following coefficients to obtain a balanced chemical equation.
C3H8
+
5 O2
→3 CO2
+
4 H2 O
a) The coefficients in equation (1:5:3:4) indicate the ratio of moles as well as molecules.
Thus,
1 mole + 5 moles →3 moles
+ 4 moles
b) 1 liter
+ 5 liters
→3 liters + 4 liters
Self-Exercises :
A.
liters.
1.
After balancing the following equations, interpret the coefficients in terms of moles and
C4H10
+
2. Pb(NO3)2
3. H2SO4
O2
→→
+ Fe2(SO4)3
+ NH4OH
CO2
→
+
H2 O
PbSO4
+ Fe(NO3)3.
→ (NH4)2SO4 + H2O
Mole- mole Problems
A type of calculation that relates the moles of two substances participating in a balanced
chemical equation.
The coefficients in the chemical reaction also indicate the ratio of moles of reactants and
products. Consider the complete combustion of natural gas. Methane, CH4, reacts with oxygen to
give carbon dioxide and water. The equation is
CH4 (g) + 2 O2 (g)
→ CO2 (g) + 2 H2O (g)
The coefficients indicate that 1 mole of methane reacts with 2 mole of oxygen. Suppose
we wanted to find out how many moles of oxygen react with 2.25 moles of CH4, we can set up the
problem as follows:
2.25 mol CH4 X
2 mol O2
1 mol CH4
=
4.50 mol O2
We could also calculate the moles of carbon dioxide produced from the reaction. In the
balanced equation, we have 1 mol CH4 yields 1 mol CO2, applying the unit factor we have
2.25 mol CH4 X
1 mol CO2 = 2.25 mol CO2
1 mol CH4
In this type of problem, we can convert from moles of moles of one substance to moles of
another using a single unit. This type of conversion is sometimes referred to as a MOLE- MOLE
PROBLEM.
Self-Exercises :
1.
Calculate the number of moles of hydrogen that react with 0.125 mol of nitrogen gas. Also
calculate the number of ammonia, NH3, produced from the reaction.
N2
+ 3 H2 → 2NH3
2.
Iron is produced in industrial blast furnace by passing carbon monoxide gas through
molten iron (IV) oxide at 15000 C .The balanced equation is
Fe2O3
+
CO → Fe
+ CO2
a)
How many moles of carbon monoxide react with 2.50 mol of iron(IV) oxide?
b)
How many mol of Iron are produced from 2.50 mol of Fe2O3.
3.
How many moles of oxygen gas react with 2.50 mol of phosphorous? How many moles of
diphosphorous pentaoxide are produced?
P
+
O2
→ Fe2O5
MASS- MASS STOICHIOMETRY PROBLEMS
A mass-mass stoichiometry is so named because an unknown mass of a substance is
calculated from a given mass of a reactant or product in a chemical equation. The process for
solving mass-mass stoichiometry problems is outlined as
Mass of
Moles of
Given
→
Moles of
Given→
Unknown
mass of
→
unknown
After balancing the chemical equation proceeds as follows :
a) convert the relevant given mass of substance into moles of using the molar mass as a unit factor
b) convert the number of moles of a given mass of a given substance into moles of unknown
substance using the coefficients in the balanced chemical equation
c) calculate the mass of the unknown substance by using the molar mass of the unknown as a unit
factor..
Example :
1. Consider the high temperature reduction of 14.4 g iron (II) oxide to elemental iron with aluminum
metal. The balanced equation is:
3 FeO
+
2 Al → 3 Fe
+ Al2O3
To calculate the mass of aluminum necessary for the reaction requires 3 steps.
a) Calculate the moles of Iron II oxide. Since the molar mass of FeO = 71.80 g/mol
14.4 g FeO X 1 mol FeO = 0.201 mol FeO
71.8 g FeO
b) Convert the number of moles to moles of Al by applying the coefficient from the balanced
chemical equation.
0.201 mol FeO x
2 mol
= 0.134 mol Al
3 mol FeO
c) Use the molar mass of Al as a unit factor to obtain mass of Al reacting with 14.4 g of FeO thus,
0.134 mol Al x
27.0 g Al = 3.61 g Al
1 mol Al
The solution to the above problem can be shown as
14.4 g FeO X 1 mol FeO
71.8 g FeO
x 2 mol Al
x
3 mol FeO
27.0 g Al
= 3.61 g Al
1 mol Al
MASS –VOLUME STOICHIOMETRY PROBLEMS
In a mass-volume stoichiometry problem, we first convert the mass of a substance to
moles, we then convert the moles of a given substance to moles of an unknown substance using
the coefficients of the balanced equation. Finally, we multiply the moles of the unknown gaseous
substance by the molar volume to obtain the volume in liters.
Mass of
Given
Moles of
→
Given
Moles of
→
Unknown
gaseous volume
→
unknown
We must always begin with a balanced chemical equation. After balancing the equation,
proceed as follows:
a)
Convert the relevant given mass of substance into moles using the molar mass as a unit
factor.
b)
Convert the number of moles of given substance into moles of unknown substance using
the coefficients in the balanced chemical equation.
c)
Calculate the volume of the gas (2.4 L/mol at STP)
It is also possible to perform the reverse calculation, that is, we can find a mass of unknown
substance given the volume of gas.
Gaseous
Volume of →
Given
Moles of
given →
Moles of
Unknown
→
mass of
Unknown