CHEMISTRY 10 UNIT 1.pptx

UNIT 1
FUNDAMENTAL LAWS OF
CHEMISTRY AND CHEMICAL
CALCULATIONS

LAW OF CONVERSATION OF MASS
• French chemist Lavosier stated that ‘atoms of an object
can not be formed or destroyed, but can be moved
around and changed into different particles’.
• This law allowed chemistry to be a respectable science
• With the advent of this law, chemists took the mystery
and illusion of alchemy and brought predictability and
reliability to the science of chemistry
• If scientists know the quantities and identities of
reactants for a particular reaction, they can predict the
amount of products to be produced

LAW OF CONVERSATION OF MASS
• Example
– When 50 grams of CaCO3 is heated, it decomposes
into CaO and CO2 gas.
– How many grams of CaO is formed based on the
fact that CO2 formed is 22 grams.
– Solution: We have 50 gram reactants. Thus, we
must have 50 gram product. The mass of CaO
formed becomes 50-22 = 28 g.
– Exercise: How many grams of hydrogen (C) are
present in 32 grams of methane (CH4) containing
24 grams of carbon (C)?

LAW OF DEFINITE PROPORTIONS
• ‘In a given chemical compound, the elements are
always combined in the same proportion by
mass’.
• Joseph Proust conducted many experiments and
found that ‘the percentage composition of a
compound by mass is always constant’.
• For instance, water will always have 2 g of
hydrogen and 16 g of oxygen.
• Thus, the ratio of hydrogen to oxygen is 2/16.
• Therefore, if we had 36 g of water, we would have
4 g of hydrogen and 32 g of oxygen.

SOLUTION
A) As the iron/sulfur ratio is 7/4, 11 g of Iron (II) sulfide is
formed when combined. However, the graph shows we
have 22 g of Iron (III) sulfide. We have to double the
amount of iron and sulfur. Thus, we must have 14 g of iron
and 8 g of sulfur.
B) Between time 0 and 2t, the elements combine. Thus,
the amount of product increases. However, at time 2t, the
reaction is completed. After this point, the amount of
product must stay constant.
C) For 11 g of Iron (III) sulfide, we need 7 g of Fe and 4 g of
S. Now that we have 16.5 g of the product (11x1.5 =16.5).
We must multiply the amount of Fe and S by 1.5. Hence,
we will have 7x1.5= 10.5 g of Fe and 4x1.5= 6 g of S.

EXERCISE
A) How many grams of sulfur are required to
convert 21 g of iron into iron sulfide?
B) How many g of each substance is needed to
have 66 g of iron (III) sulfide?

EXAMPLE
• In one experiment, 2 g of Na metal reacted with
chlorine gas, yielding 5.08 g of NaCl. In another
experiment, 2 g of chlorine gas reacted with Na,
yielding 3.3 g of NaCl. Prove that these results are
consistent with the law of definite proportions.
•First experiment Na/Cl2 ratio is (2/3.08 = 0.65)
•Second experiment Na /Cl2 ratio is (1.3/2 = 0.65)
•Therefore, we conclude that the substances
combine in the same ratio.

EXAMPLE
• The ratio of A to B by mass in a compound of A
and B elements is 6/15.
•What is the maximum amount of compound
produced when 20 g of A reacts with 45 g of B?
•Let us divide 20/45, we get 6.667/15. Therefore,
we have some extra g of element A.
• With 45 g of B (15 x3), 18 g of A (6 x3) can react.
• Thus, 45+18 =63 g of product with 2 g
of A remaining excess.

EXERCISES
• In a compound, the ratio of A to B by mass is
2/5. If 20 g of each element is reacted to give
a reaction, how many grams of the product
(compound) can be obtained at most?
• One-fifth of a compound containing X and Y is
X. When 20 g of a mixture of X and Y is
reacted, 5 g of X remains unreacted. What is
the weight of X in the mixture?

EXAMPLE
• In the compound X2Y3, the mass ratio of X to Y is 9/10.
• If we have 20 g of each of X and Y,
• A) Which element remains behind, and how many g?
• B) How many g of X2Y3 is produced?
• A) 9 g of A reacts with 10 g of B. Thus, 18 g of A can react
with 20 g of B, indicating that we will have 2 g of A in excess.
• B) We will have 18+20=38 g of product.
• EXERCISE
• The compound CaBr2 is 20% Ca by mass. What is the
combining ratio between calcium and bromine by mass
(Ca=40)
• Solution: If 20% of the compound is Ca, then the remaining
80% must be Br2.
• Therefore, 20/80 = 1/4.

LAW OF MULTIPLE PROPORTIONS
• ‘When two elements combine to form more
than one compound, the different masses of
one that combine with the same mass of the
other are in the ratio of small whole numbers’.
• Dalton and other scientists posited this
theory.
• Let us see an example mc mo (mo)CO/(mo)CO2
• Carbon monoxide (CO) 12 16 16/32= 1/2
• Carbon dioxide (CO2) 12 32

LAW OF MULTIPLE PROPORTIONS
Let us assume that in one of the compounds, 14 g of
nitrogen has combined with 16 g of oxygen. Yet, in
another compound, 14 g of nitrogen reacted with 40 g
of oxygen.
The masses of oxygen that combine with 14 g of
nitrogen are in the ratio of 16/40 = 2/5.These numbers
are always integers (not fractions) because atoms
cannot be divided.
Hence, the law of multiple proportions suggests that a
chemical element comprises small, indivisible,
indestructible particles called atoms, and all atoms of a
given element are identical in mass (weight).

EXAMPLE
Manganese and oxygen yield two different compounds, Mn2O3
and Mn3O4. In Mn2O3,
110 g Mn combines with 48 g of O2, while in Mn3O4, 165 g Mn
reacts with 64 g of O2.
What is the ratio of oxygen atoms by mass combined with the
same amount of Mn?
a) We must have the same amount of Mn. Therefore, we must
equate the number of Mn atoms. We multiply 2 by 3 and 3 by 2.
Then, we have 330 g of Mn in each case.
The ratio of oxygen then becomes 48x3/64x2 =9/8
b) Instead of weights, we can use the number of atoms.
To have the same number of Mn atoms, we multiply by 3 and 2.
Then, we also need to multiply by 3 and 2 for oxygen. Thus, 3x3
gives 9, and 2x4 gives 8.

EXERCISE
• X and Y form two different compounds. In the
first compound, 24 g of X combines with 6 g of
Y. 60% by mass of the second compound is X.
What are the multiple proportions in these
compounds?
• mx/my = x/y= 24/6 =4 in the first compound
• In the second compound, mx/my= 60/40 =1.5
• Thus, multiple proportion is 1.5/4 = 3/8.

EXAMPLE
•The elements X and Y combine to give two different
compounds.
• 7 g of X and 2 g of Y combine to produce XY.
• How many g of Y combines with 7 g of X to yield
X2Y3?
• In compound XY, mx/my= x/y =7/2
• In compound X2Y3, mx/my = 2.X/3.Y = 2x7/3x2
•mx/my= 7/3
•Therefore, 7 g of X combines with 3 g of Y to yield
10 g of X2Y3.

MOLE CONCEPT
• The smallest unit of an element is an atom, and the smallest
unit of a compound is a molecule.
• 1 g of Hydogen contains 6.02 x 1023 hydrogen atoms.
• In 1961, The International Union of Pure and Applied Chemistry
(IUPAC) adopted to choose Carbon-12 (Carbon twelve) as
standard to compare the masses of atoms.
• Atomic mass of C is 12 amu.
• Thus, 1/12 of the mass of C-12 is 1 atomic mass unit (amu).
• One sulfur atom is twice heavier than an oxygen atom.
• One hydrogen atom is 12 times lighter than a carbon atom.

MOLE AND AVOGADRO’S NUMBER
• Carbon—12 isotope is the reference.
• This number, 6.02 x 1023, is called Avogadro’s
number.
• Thus, a mole contains 6.02 x 1023 items (atoms,
molecules, eletrons, or other entities).
• 1 mol = 6.02 x 1023 entities.
• If a sample of an element contains N atoms, the
number of moles of the element (n) can be found
by dividing N by Avogadro’s number, NA.
• n (mol) = N (particles) / NA (particles/mol)
• NA= 6.02 x 1023

EXAMPLE
The population of the world is about 8 billion.
How many moles of people are there in the world?
6.02 x 1023 people 1 mol
8 x 109 people x mol
X= (8 x 109)/(6.02 x 1023) = 1.3 x 10-14 mol.
EXERCISE
An average breath comprises about 0.09 mole molecules. What is the number of
molecules in this air sample?
EXERCISE
For an experiment, two moles of iron atoms are required.
How many iron atoms are required for this experiment?
Because we have two moles, the number of atoms is 2x 6.02 x 1023 or 1.204 x 1024.
EXERCISE
An average signature written with a pencil contains about 8 x 10-5 mol of carbon
atoms. How many carbon atoms are used for the signature?
EXERCISE
We have 0.1 mol (1.8 g) of water in a container. If one billion water molecules
evaporate in a second, how many years will it take to evaporate all the water in the
container?

MOLAR MASS
• Atomic Mass or Atomic Weight
The mass in grams of an element's one-
mole atom is called the element's molar weight.
The unit of molar weight is g/mole and is usually
abbreviated as MW.
An oxygen atom has a mass of 16 amu.
One mole oxygen atom weighs 16 g
The atomic weight of oxygen is 16.

MOLAR MASS
6.02 X 1023 atoms 12 g
1 ‘C’ atom = 12/6.02 x 1023
1 amu = (1/12) C atoms
1 amu = (12/6.02 x 1023) x(1/12) = 1.67 x 10-24 g.
Thus, 1 amu= 1.67 x 10-24 g
Remember that an element's molar mass does not
represent the absolute weight of an individual atom
but the total mass of 6.02 x 1023 atoms of that
element.

MOLAR MASS
• The atomic weight of iron atom is 56.
• Therefore, 1 atom weighs 56/6.02x 1023 g.
• We may get the same result using the
information 1 amu = 1.67 x 10-24 g.
• Then, one iron atom has a relative mass of 56
amu, the actual mass of iron atom is 56 x
1.67x10-24 g or 9.3 x 10-23 g.
• n (mol) = (m (g) /AW (g/mol))

EXAMPLE
• How many moles of calcium atoms are there in 8 g of
calcium? How many calcium atoms? (The atomic weight of
calcium is 40.)
• 40 g of calcium 1 mol
• 8 g of calcium x mol x= 0.2 mole calcium
• 1 mol 6.02 x 1023 atoms
• 0.2 mol x atoms x = 1.2 x 1023
• Exercise: What is the number of moles of carbon atoms in 18
g of carbon? How many carbons are there in 18 g of carbon?
• Number of moles = 18 g of carbon/12 g/mol = 1.5 mol
• 1 mol 6.02 x 1023 atoms
• 1.5 mol x atoms X= 9.03 x 1023

EXAMPLE
• If 1000 atoms of an element x weighs 9.3 x 10-20 g,
find the molar weight of element x.
• For any element, the weight of 6.02 x1023 atoms is
the molar weight.
• 1000 atoms 9.3 x 10-20 g
• 6.02 x1023 x g x=56 (Fe)
• EXERCISE
• A) How many copper atoms are in 1.28 g of copper?
• B) What is the mass of one copper atom?
• (The atomic weight of copper is 64.)
• Let us also do the exercise on page 26.

Molecular Mass or Molecular Weight
• 1 mol H2O = 6.02x1023 H2O molecules =18 g.
• Thus, MWH2O = 18 g/mol.
• 1 mol CO2 = 6.02x1023 CO2 molecules = 44 g.
• Thus, MW CO2 = 18 g/mol.
• 1 mol of H2O contains 2 mols of H atoms (2 g), and 1
mol of O atom (16 g).
• The molecular weight of H2SO4 (H: 1, S: 32, O : 16)
• MW H2SO4 = (2x1) + (1x32) + (4x16) = 98 g/mol.
• The molecular weight of Fe2 (SO4)3 (Fe:56, S:32, O:16)
• MW Fe2 (SO4)3= (2x56) + (3x32) + (12x16) = 400 g/mol.

EXAMPLE
• Calculate the molecular weights of each of the
following compounds.
• a) SO3 b) Fe2O3 c) Cu (NO3)2 d)Ca3(PO4)2 e)
CuSO4.5H2O
• a) (1x32) + (3x16) = 80 g/mol
• b) (2x56) + (3x16) = 160 g/mol
• c) (1x64) + (2x14) + (6x16)=188 g/mol
• d) (3x40) + (2x31) + (8x16) = 310 g/mol
• e) (1x64) + (1x32) + (4x16) + (5x18) = 250 g/mol

EXERCISE
• Calculate the molecular weights of each of the
following compounds.
• A) P2O3 b)C2H6 c) Al(OH)3 d) (NH4)2SO4
e) Na2SO4.10H2O
The number of moles of m grams of compound
with a molecular weight of M can be found with
the following formula:
n(mol) = m(g) / MW(g/mol)

EXAMPLE
• Answer these questions for butane, C4H10, a component of the house-
hold gas used for cooking in kitchens (C:12, H:1).
• A) What is the mass of one mol of butane?
• B) What is the mass of 0.25 mol of butane?
• C) What is the number of molecules in 2.9 g of butane?
• D) What is the mass of 1.5 x 1022 butane molecules?
• E) What is the mass of a single molecule of butane?
• A) (4x12) + (10x1) = 58 g/mol
• B) 0.25 mol x 58 g/mol = 14.5 g
• C) 2.9 g butane/58 g butane/mol = 0.05 mol
• 1 mol 6.02 x 1023 molecules
• 0.05 mol x molecules X= 3.01 x 1022 molecules
• D) 6.02 x 1023 molecules 1 mol = 58 g
• 1.5 x 1023 molecules x x=0.025 mol= 0.025 x58 = 1.45 g
• E) 6.02 x 1023 molecules 58 g
• 1 molecule x g x= 9.63 x 10-23 g.

EXERCISE
• 0.1 mol of compound XO weighs 3 grams. If one compound
molecule has 15 neutrons, what will be the group and the
period numbers of element X in the periodic table? (16 O
8)
• 0.1 mol compound 3 g
• 1 mol compound L g L= 30 g
• The compound has 15 neutrons. Thus, it must have (30-15
=15) protons.
• P = e, Thus, it must contain 15 electrons.
• O has 8 electrons. Hence, X has (15-8 = 7 electrons).
• 7 X = ) )
2 5 Thus, 2nd period, group VA.

EXERCISE
• 37.5 g of Al2X3 contains 0.5 mol of aluminum.
(Al:27)
• A) What is the molecular weight of the
compound?
• B) What is the atomic weight of element X?
• C) If the number of protons and electrons are
the same in the nucleus of element X, what
will be the place of element in the periodic
table? (O:16)

MOLAR VOLUME
The volume occupied by one mole of any substance is called its molar volume.
1 mol Al = 27 g/mol d Al = 2.7 g/cm3 V= m/d
27 (g/mol) / (2.7 g/cm3) = 10 cm3/mol
1 mol NaCl = 58.5 g/mol d Nacl = 2.2 g/cm3
V= m/d 58.5 (g/mol) / (2.2 g/cm3) = 26.6 cm3/mol
1 mol H2O = 18 g/mol d H2O = 1 g/cm3
V= m/d 18 (g/mol) / (1 g/cm3) = 18 cm3/mol
Solid nitrogen
1 mol N2 = 28 g/mol, d= 1.03 g/cm3 (below -210C)
V= (28 g/mol)/(1.03 g/cm3) = 27.2 cm3/mol
Liquid nitrogen
1 mol N2 = 28 g/mol, d= 0.81 g/cm3 (at -210C)
V= (28 g/mol)/(0.81 g/cm3) = 34.6 cm3/mol
Gaseous nitrogen
1 mol N2 = 28 g/mol, d= 0.00125 g/cm3 (at OC and 1 atm)
V= (28 g/mol)/(0.000125 g/cm3) = 22.4x10 3 cm3/mol or 22.4 L

MOLAR VOLUME
• Even though the molar volumes of solids and
liquids change from substance to substance, the
molar volumes of all gases are the same under
the same conditions of temperature and
pressure (usually 0C and 1 atm).
• These standard temperature and pressure
conditions are abbreviated as STP. One mole of
any gas occupies a volume of 22.4 L at STP.
• 1 mol H2 molecules= 6.02x10 23 H2 molecules = 2 g H2 = 22.4 L
• 1 mol CO2 molecules= 6.02x10 23 CO2 molecules = 44 g CO2 = 22.4 L
• 1 mol NH3 molecules= 6.02x10 23 NH3 molecules = 28 g NH3 = 22.4 L

Example
• Calculate the volume of each of the following
gases at STP. (C:12, O:16)
a) 0.25 mol O2 b) 17.6 g CO2 c) 3.01x1022 HF molecules
Solution a) 0.25 mol O2 x 22.4 L/mol O2 = 5.6 L
b) (17.6 g CO2/44 g CO2/mol) x 22.4 L/mol = 8.96 L
c) 3.01x1022 HF molecules x (1 mol HF/6.02x1023 HF
molecules) = 0.05 mol HF
(0.05 mol HF) x(22.4 L/1 mol HF) = 1.12 L.

EXERCISE
• Answer the following questions for 6.72 L of SO2
gas at STP. (S:32, O:16)
• A) What is the number of moles?
• B) How many molecules are in it?
• C) How many moles of sulfur atoms are in it?
• D) How many oxygen atoms are in it?
• E) How much does it weigh?
• F) How many grams of oxygen does it have?
• G) How many atoms are contained?

ISOTOPES AND MOLAR MASS
•Atoms having the same number of protons (atomic number)
but different numbers of neutrons (thus atomic mass number)
are called isotopes. It is customary to identify isotopes by mass
number only, such as O-16, O-17, and O-18.
•To calculate the atomic weights of elements having isotopes,
we use the weighted average concept.
•Average atomic weight= Summation (fractional abundance x
Atomic mass)•Example: Natural magnesium is found to
contain 78.6% Mg-24, 10.1% Mg-25 and 11.3% Mg-26.
Calculate the weighted average of magnesium atoms.
•Solution: (0.786x24 + 0.101x25 + 0.113x26) = 24.33.
•EXERCISE: Naturally occurring boron comprises 80.4% B-11
and 19.6% B-10. Calculate the average atomic mass of boron.

CHEMICAL REACTIONS
• Chemical reactions occur when chemical bonds between
atoms are formed or broken.
• The substances that go into a chemical reaction are called
reactants, the substances procudes are termed as the
products.
• Physical appearances of products often differ from
reactants.
• Chemical reactions are often accompanied by the
appearance of gas, fire, precipitate, color, light, sound, or
odor.
• For instance, the oxidation of propane releases heat and
light, and a rapid reaction is an explosion
• C3H8 + 5O2→ 3CO2 + 4H2O

In chemical reactions
A) Mass is conserved. The mass of reactants equals the
mass of the products.
2H20→ 2H2 + O2
(36 g= 4 g + 32 g)
B) Atoms are conserved. Each type's total number of
atoms remains unchanged in a chemical reaction.
2H20→ 2H2O2 (4 H atoms ↔ 4 H atoms, 2 O atoms ↔ 2 O
atoms)
C) The chemical bonding changes: The arrangement of atoms
differs between reactants and products (2H-O-H)→ 2H-H + O=O.
D) The number of mole, molecules, and volume of the substances
may not be conserved in chemical reactions.
E) The electrical charge is conserved: The total charge of reactants
must be the same as that of products.

In writing a chemical reaction
• A) We must know what the reactants and
products are.
• B) We must write the correct formula for each
reactant and product.
• C) We must show that atoms are conserved.
Propane + Oxygen→Carbon dioxide + Water (in words)
C3H8 + O2→ CO2 + H2O (skeleton equation).
We must balance the equation to satisfy the
conservation of mass.
The process of finding the coefficents is called balancing
equation.

BALANCING AN EQUATION (STEPS)
Step 1. Fix the substance's coefficient with the largest number
of atoms per molecule as 1.
In our case, C3H8 must have a coefficient of 1.
Step 2. Look for the element(s) in the most complex substance
appearing only once on each side of the equation.
In our case, two elements (C and H) appear only once on each
side. Thus, we must put 3 in front of CO2. Then, to balance H,
we must place 4 in front of H2O. As a last step, we can balance
O atoms. We must put 5 in front of O2.
Check
We have 3 C atoms on each side, 8 H atoms on each side, and
10 O atoms on each side.
Then, the answer is C3H8 + 5O2→ 3CO2 + 4H2O.

EXAMPLE
In the lab, oxygen gas may be generated by decomposing
potassium chlorate into potassium chloride and oxygen
gas. Write a balanced equation describing the reaction.
Potassium chlorate→ Potassium chloride + Oxygen gas
KClO3→ KCl + O2
1KClO3 →KCl + O2
C is already balanced; we can put 3/2 in front of O2 on the right
side.
KClO3→ KCl + 3/2O2
Even though we may see fractional coefficients in chemistry, it is
more common to have whole numbers. Hence, we can multiply
each coefficient by 2.
The final balanced equation becomes 2KClO3→ 2KCl + 3O2

EXERCISE
• Balance each of the following equations:
• A) CH4 + Cl2 → CCl4 + HCl
• B) Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O
• C) Fe + H2O → Fe3O4 + H2
• D) Al4C3 + HCl → AlCl3 + CH4

EXERCISE
• Determine the formula of x in the following
balanced equations.
• A) 9Fe2O3 + 2NH3 → 6X + N2 + 3H2O
• B) 2MnO2 + 4x + O2 → 2K2MnO4 + 2H2O
• C) 2KMnO4 + 16HCl → 2MnCl2 + 2KCl + 5x + 8H2O
• D) Cr2O3 + 3X → 2CrCl3 + 3CCl2O
• E) As2S3 + 12NaNO3 + 3 Na2CO3 → X + 3Na2SO4
+ 3CO2 + 12NaNO2

TYPES OF CHEMICAL REACTIONS
• Combustion Reactions
• Combination Reactions
• Decomposition Reactions
• Acid-base Reactions
• Dissolution-precipitation Reactions

1) Combustion Reactions
• In combustion reactions, carbon-hydrogen compounds
(hydrocarbons) or carbon-hydrogen-oxygen
compounds burn completely in oxygen gas to produce
carbon dioxide and water.
• C2H6 (g) + 7/2O2 → 2CO2 + 3H2O (g) + heat
• C2H5OH (l) + 3O2 → 2CO2 + 3H2O (g) + heat
• In some combustion reactions, heat and light energy
are produced, while some are flameless. Hence, the
light is not released. The burning of wood, coal, and
candle are fast; iron corrosion and copper oxidation are
slow-burning phenomena.

2) Combination (Synthesis) Reactions
• Two or more substances (elements or
compounds) react to form a more complex
substance.
• The reactants may be elements or compounds,
but the resulting product is always a compound.
• A + B → AB
• C (s) + O2(g) → CO2 (g)
• Fe (s) + S (s) → FeS (s)
• H2 (g) + Cl2(g) → 2HCl (g)
• Synthesis reactions can occur naturally or
artificially.

3) Decomposition Reactions
• In these reactions, a substance is broken down
into two or more simpler substances.
• AB → A + B
• 2KClO3→ 2KCl + 3O2
• 2HgO (s) → 2Hg (l) + O2 (g)

4) Acid-Base Reactions (Neutralization)
• In aqueous solutions, acids increase the
hydrogen ion (H+)
• HCl (aq) → H+ (aq) + Cl- (aq)
• A base is a compund dissolving in water to
give hydroxide ions, OH-.
• NaOH (aq) → Na+ (aq) + OH- (aq)
• When an acid and base react in an aqueous
solution, H+ and OH- ions combine to form
water. These ions ‘neutralize’ one another.

Selected Acids and Bases
• ACID FORMULA BASE FORMULA
Hydrochloric HCl Sodium NaOH
Acid Hydroxide
Hydrobromic HBr Potassium KOH
Acid Hydroxide
Hydroiodic HI Magnesium Mg(OH)2
Acid Hydroxide
Nitric HNO3 Calcium Ca(OH)2
Acid Hydroxide
Sulfuric Acid H2SO4 Barium Ba(OH)2
Hydroxide
Phosphoric Acid H3PO4 Ammonia NH3

Acid-Base Reactions (Neutralization)
• Acid + Base → Salt + Water
H2SO4(aq) + 2KOH (aq) → K2SO4 (aq) + 2H2O (l)
sulfuric acid potassium hydroxide potassium sulfate water
H2SO4(aq) + Mg(OH)2 (aq) → MgSO4 (aq) + 2H2O (l)
sulfuric acid magnesium hydroxide magnesium sulfate water
2HNO3(aq) + Ca(OH)2 (aq) → Ca(NO3)2 (aq) + 2H2O (l)
nitric acid calcium hydroxide calcium nitrate water

5) Dissolution-Precipitation Reactions
• A precipitation reaction is a reaction where two or more
solutions are combined to yield an insoluble product, a
precipitate.
• The interactions between two ions produce an insoluble solid
product or a precipitate.
• The reaction between lead (III) nitrate (Pb(NO3)2) and
potassium iodide (KI) is unique.
• A yellow precipitate PbI2 is formed. The potassium and nitrate
ions remain in the solution. We can write ionic and net ionic
equations for the reaction between Pb(NO3)2 and KI.
• Pb2+ (aq) + 2NO3- (aq) + 2K+ (aq) + 2I-(aq) → 2K+ (aq) + 2NO3-
(aq) + PbI2(s)
• The nitrate and potassium ions have the same form on each
side of the equation, so they are eliminated as spectator ions.
Then, the net equation is as follows.
• Pb2+ (aq) + 2I-(aq) → PbI2(s)

CALCULATIONS BASED ON CHEMICAL EQUATIONS
2Al (s) + 6HCl (g) → 2AlCl3 (s) + 3H2 (g)
2 atoms 6 molecules 3 molecules 3 molecules
2 mol 6 mol 2 mol 3 mol
2x27 g 6 x36.5 g 2 x 133.5 g 3 x 2 g
---- 6 x 22.4 L 3 x 22.4 L -------
As you can see, the volume is only for gases. The
coefficients in a balanced chemical equation indicate
the number of molecules or moles of molecules (or
atoms for elements) of substances that react or are
formed.

Calculations About Quantities of
Reactants and Products
• STEP I: Write the balanced equation
• STEP II: Convert the quantity of given
substance into moles.
• STEP III: Find the mole of desired substance by
using the relationship between coefficients of
given substance and desired substance in the
balanced equation.
• STEP IV: Convert the moles of desired
substance into the desired units.

EXAMPLE
When mercuric oxide, HgO, is heated, it releases mercury and oxygen
gas. How many moles of oxygen gas will be produced from the
decomposition of 4 moles of mercuric oxide?
Step 1: 2HgO →2Hg + O2
We have 4 moles of HgO
2 moles HgO 1 mol O2
4 moles HgO x mol O2 X= 2 moles of O2
How many potassium chlorate molecules are required to prepare 6
moles of oxygen gas?
2KClO3→ 2KCl + 3 O2
2 moles of potassium chlorate 3 moles of O2
X moles of potassium chlorate 6 moles of O2 x= 4 moles
1 mol 6.02 x 1023 molecules
4 moles x x= 2.408 x 1024 KClO3 molecules

EXERCISE
• When aluminum is heated with sulfur, aluminum sulfide is formed.
a) Write the balanced equation?
b) Calculate the number of Al moles combining with 6 moles of S.
c) Calculate the number of Aluminum sulfide molecules produced
from 0.4 mol of Al.
Solution
a) 2Al + 3S → Al2S3
b) 3 moles of S 2 moles of Al
6 moles of S x moles x= 4 moles of Al
c) 2 moles of Al 1 mol of Al2S3
0.4 moles of Al x x=0.2 moles of Al2S3
1 mol 6.02 x 1023 molecules
0.2 mol x x= 1.2 x 1023 molecules.

EXAMPLE
How much sodium hydroxide is formed if 212 g of sodium bicarbonate
reacts with calcium hydroxide?
Na2CO3 + Ca(OH)2 → 2 NaOH + CaCO3 (Na2CO3 = 106 g/mol; NaOH =40
g/mol)
nNa2CO3 = 212 g Na2CO3/106 g/mol = 2 moles
1 mol of Na2CO3 2 mol NaOH
2 moles of Na2CO3 X X = 4 moles of NaOH
Or we can use the weights
106 g Na2CO3 80 g NaOH
212 g Na2CO3 X g NaOH x= 160 g NaOH
Exercise
How much sodium hydroxide will react with 49 g of phosphoric acid?
(H3PO4 = 98 g/mol, NaOH =40 g/mol. The equation follows.
(H3PO4 + 3 NaOH→ 3H2O + Na3PO4.

EXAMPLE
• How many liters of carbon dioxide form from the
combustion of 60 liters ethylene gas (C2H4)?
• Let us recall combustion of an compound
containing C and H must yield carbon dioxide and
water vapor.
• C2H4 + 3O2→ 2CO2 + 2H2O
• 1 mol C2H4 2 mol CO2
• 1 volume C2H4 2 volumes of CO2
• 1 liters 2 liters
• 60 liters x liters x= 120 liters.

EXERCISE
Given the balanced equation:
N2 (g) + 3H2 (g) → 2NH3 (g), Calculate.
a) The volume of H2 reacting with 12 L of N2
b) The volume of NH3 produced from 4 L of N2
c) The volumes of N2 and H2 to produce 60 L of
NH3. Assume that all volume measurements
are made under identical conditions.

EXAMPLE
Aluminum reacts with hydrochloric acid and produces aluminum chloride and
hydrogen gas.
Answer the following questions for 2.7 g of aluminum used in the reaction.
(Al:27, Cl:35.5, H:1)
A) How much AlCl3 will be produced?
B) How many moles of HCl are used?
C) How many H2 molecules are formed?
D) What is the volume of H2 at STP?
• Solution: The balanced equation : 2Al + 6HCl→ 2AlCl3 + 3H2
A) n(Al) = 2.7 g/(27 g/mol) = 0.1 mol.
2 mol Al 2 mol AlCl3 produced
0.1 mol Al x x= 0.1 mol = 0.1 x(133.5g) = 13.35 g
B) 2 moles Al 6 moles HCl
0.1 mole Al x x= 0.3 mol HCl is utilized.
C) 2 mol Al 3 x 6.02 x1023 H2 molecules
0.1 mol Al x x= 9.03 x 1022 H2 molecules
d) 2 moles Al 3 x 22.4 L H2
0.1 mole Al x x= 3.36 L H2.

Reactions Involving Limiting Reactants
• In some cases, all reactants may not be consumed
completely.
• If so, the reaction continues until one reactant is used
up.
• The reactant that is consumed first is the limiting
reactant.
• The reactant that is partially consumed or partially
unreacted is the excess reactant.
•The limiting reactant controls the amount of product
that can be formed.
• Hence, while solving problems involving excess
reactants, the limiting reactant is identified first, and the
amount of product produced by the reaction is based on
the quantity of the limiting reactant.

EXAMPLE
The mixtures of 2 moles of H2 and 2 moles of O2 gases are ignited to yield water.
Which reactant is limiting? Which reactant is excess? How much water is
produced?
Let us write the balanced equation first.
2 H2 (g) + O2 (g) → 2H2O (l)
2 moles 1 mole
2 moles of H2 can react with 1 mol of O2. However, we have 2 moles of O2,
suggesting that O2 is the excess and H2 is the limiting reactant.
2 moles of hydrogen gas 2 moles of water
2 moles of hydrogen gas x X= 2 moles of water.
Therefore, we will have 2 moles or 2 x18 = 36 g of water.
Another approach:
The ratio in the equation: H2/O2 = 2/1
The ratio in the real condition: H2/O2 = 2/2
Hence, O2 is the excess reactant.

EXAMPLE
Carborundum, SiC, is used as an abrasive. It is formed by
combining SiO2 and carbon, based on the reaction: SiO2 + 3C →
SiC + 2CO
How much SiC is formed from 6 g of SiO2 and 6 g of C?
Mol (n) of SiO2 = 6 g/60 g/mol = 0.1 mol
Mol (n) of C = 6 g/12g/mol = 0.5 mol
1 mol of SiO2 3 moles of C
0.1 mol of SiO2 x x= 0.3 mol of C.
Because we have 0.5 moles of C, 0.2 mol of C will be excess.
Thus, the limiting reactant is SiO2.
1 mol of SiO2 yields 1 mol of SiC
0.1 mol of SiO2 x x= 0.1 mol of SiC.
Therefore, 0.1 mol of SiC x 40 g SiC/mol = 4 g of SiC is formed.

EXERCISE
• How much water is produced when 4 g of H2 and 8 g of O2 are
reacted?
• Example
– Given : N2 (g) + 3H2 (g) → 2NH3
– If 20 L of N2 and 15 L of H2 gases at the same conditions are
reacted, which gas will be excess and how many liters?
– 1 mol of N2 3 moles of H2
– 1L of N2 3 L of H2
– 20 L of N2 x x= 60 L of H2. However, we only have 15
L of H2 gas, making H2 the limiting reactant.
– 1 L of N2 3 L of H2
– x 15 L of H2 x= 5 liter. The excess reactant
is N2. We will have 15 L of N2 in excess.

EXERCISE
• Given the reaction 2H2S (g) + 3 O2 (g) → 2H2O (g) + 2SO2
(g). For each case, determine the excess reactant and the
amount of water produced.
• a) 2 moles of H2S and 4 moles of O2
• b) 4 moles of H2S and 6 moles of O2
• c) 8.5 g of H2S and 9.6 g of O2
• d) 6.02 x 1022 H2S molecules and 6.72 L of O2 at STP.
• e) 4.48 L of H2S at STP and 19.2 g of O2
• f) 17.92 L of H2S at STP and 13.44 L of O2 at STP.
• EXERCISE
• Nitric oxide, NO, is prepared according to the following
reaction in labs. 3KNO2 + KNO3 + Cr2O3→ 4NO (g) +
2K2CrO4. (K:39, N:14, Cl:35.5)
• Calculate the maximum volume of NO at STP when we
have 100 g of each reactant.

EXAMPLE
• Equal amounts sodium and chlorine gas are reacted in a
closed container to yield 0.1 mol of sodium chloride.
Which element will remain behind, and how much?
(Na:23, Cl:35.5)
• 2Na + Cl2 → 2NaCl
• 2 moles 1 mol 2 moles
• X1= 0.1 mol X2=0.05 mol 0.1 mol
• We need 0.1 mol Na (0.1x23 = 2.3 g) and 0.05 mol Cl2
(0.05 x 71 =3.55 g). We will accept them as 3.55 g each at
the beginning (always equal to the larger one). Therefore,
all the chlorine gas is used up. We will have 3.55-2.3 g
=1.25 g of Na unreacted.

Calculations for Substances Having
Some Impurities
Based on the following reaction, how much HCl is produced when
sulfuric acid reacts with 200 g of 87.75% pure sodium chloride?
H2SO4 + 2NaCl → Na2SO4 + 2HCl ( assume that 12.25% of the sample is
inert in the reaction, NaCl:58.5 g/mol, HCl:36.5 g/mol).
We have 200 x 0.8775 = 175.5 g pure NaCl.
Mol of NaCl (n) = 175.5/58.5 = 3 moles
2 moles of NaCl yields 2 moles of HCl
3 moles of NaCl x
x= 3 moles = 3 moles x 36.5 g/mol=109.5 g HCl.
EXERCISE
How many liters of oxygen gas can be liberated from 1700 g of solution
containing 3% hydrogen peroxide by weight?
The equation is 2H2O2 (l) → 2H2O (l) + O2 (g)

EXAMPLE
• How much H2SO4 can be produced from 300 g of FeS2 with 60% purity
(by weight)? (FeS2= 120 g/ mol, H2SO4 =98g/mol).
• 300 g x 0.60 = 180 g pure FeS2
• Mole of FeS2 = 180/120 g/mol= 1.5 moles
• 1 mol FeS2 yields 2 moles of H2SO4
• 1.5 moles FeS2 x x= 3 moles
• mH2SO4 = 3 moles x 98 g/moles = 294 g.
EXERCISE
When an unkown quantity of magnesium is reacted with HCl, 4.48 L of H2
at STP together with some magnesium chloride is formed (Mg: 24, Cl:
35.5)
a) How much magnesium chloride is formed? b) If magnesium is 96%
pure, how much Mg is used in the reaction? c) How much HCl solution
having 30% HCl by weight is needed to yield 4.48 L of H2 at STP?

% Effficiency Calculations Based on
Reaction Equations
• % efficiency = (real efficiency/theoretical efficiency) x
100
• Example:
• When 16 g of sulfur is burned, 8.96 liters of SO2 is
obtained at STP. What is the reaction efficiency?
• Solution
• S + O2 → SO2
• 1 mol 1 mol 1 mol
• 16/32 mol x mol x= 0.5 mol. However, 8.96 L;
that is, 8.96 lt/22.4lt/mol = 0.4 mol is produced.
• Percent efficiency = (0.4/0.5) x 100 = 80%.

CHEMISTRY 10 UNIT 1.pptx

Recommended

Recommended

More Related Content

Similar to CHEMISTRY 10 UNIT 1.pptx

Similar to CHEMISTRY 10 UNIT 1.pptx (20)

Recently uploaded

Recently uploaded (20)

CHEMISTRY 10 UNIT 1.pptx