Balancing Chemical Equations Chemistry SSS 1 Second Term Lesson Note for Week Three

Second Term Lesson Note for Week Three

Class : SSS One

Subject : Chemistry

Topic : Balancing Chemical Equations.

Duration : 80 Minutes

Period : Double Periods

Reference Book :

• Lagos State Unified schemes of work for Senior Secondary Schools.
• Chemistry for Senior Secondary School, SSS 1 – 3.
• Online Resources

Instructional Material : Chart showing balancing of chemical Equations.

 

Learning Objectives : By the end of the lesson learners will be able to :

i. Describe chemical Equations and explains the concept of reactant and products in equation.

ii. Explain the basic laws

iii. Explain the term stoichiometry

iv. Balancing of chemical equation.

 

Content :

Balancing Chemical Equations

Atoms are neither created nor destroyed during any chemical reaction. Chemical changes merely rearrange the atoms.
The statement above is supported by :

• Law of conservation of mass/matter
• Law of definite proportions
• Law of multiple proportions

Chemical reactions are represented in a concise way by chemical equations.

2 H2 + O2 → 2 H2O

The reacting substances, called reactants, are located on the left side of the arrow.

The substances formed, called products, are located on the right side of the arrow.

In a chemical equation, the + sign is read as “reacts with” and the arrow is read as “produces”.

Numbers in front of the formulas are coefficients, indicating the relative number molecules or ions of each kind involved in the reaction.

Coefficients of 1 are never written – they are understood.

Numbers to the lower right of chemical symbols in a formula are subscripts, indicating the specific number of atoms of the element found in the substance.
Subscripts of 1 are never written – they are understood.

A chemical equation must have the same number of atoms of each element on both sides of the arrow. When this condition is met, the equation is said to be balanced.
To count atoms, multiply the formula’s coefficient by each symbol’s subscript.

For example : 2Al₂(SO₄)₃
For Al – coefficient of 2, times subscript of 2 = 4 Aluminum atoms

For S – coefficient of 2, times subscript inside parenthesis of 1, times subscript outside parenthesis of 3 = 6 sulfur atoms

For O – coefficient of 2, times subscript inside parenthesis of 4, times subscript outside parenthesis of 3 = 24 oxygen atoms

The order in which the following steps are performed is important. While shortcuts are possible, (and you will learn about one), following these steps in order is the best way to be sure you are correct.

Balance equations “by inspection” with these steps :
Check for diatomic molecules.

Balance the metals (not Hydrogen).

Balance the nonmetals (not Oxygen).

Balance oxygen.

Balance hydrogen.

The equation should now be balanced, but recount all atoms to be sure.

Reduce coefficients (if needed). ALL coefficients must be reducable before you can reduce. An equation is not properly balanced if the coefficients are not written in their lowest whole-number ratio.

HINT: NEVER change subscripts to balance equations.

The physical state of each substance in a reaction may be shown in an equation by placing the following symbols to the right of the formula :

i. (g)   for gas

ii. (l)   for liquid

iii. (s)   for solid

iv. (aq)   for aqueous (water) solution

Stoichiometry is the quantitative study of chemical changes.

The most common type of stoichiometry calculation is a mass-mass problem.

Generally, a mass-mass problem looks like this: “given this amount of reactant, how much product will form?”

Steps in solving a mass-mass problem :

• Write a balanced equation for the reaction.
• Write the given mass on a factor-label form.
• Convert mass of reactant to moles of reactant.
• Convert moles of reactant to moles of product.
• Convert moles of product to grams of product.
• Pick up the calculator and do the math.

Mass-Mass Sample Problem :

If iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as coal burns. If a furnace burns an amount of coal containing 125 g of FeS₂, how much SO₂ (an air pollutant) is produced?

1. Write a balanced equation showing the formation of iron (III) oxide and sulfur dioxide.
4 FeS₂ + 11 O₂ → 2 Fe₂O₃ + 8 SO₂

2. Write the mass information given in the problem.

3. Convert grams of FeS₂ to moles of FeS₂.

4. Changes moles of FeS₂ (reactant) to moles of SO₂ (product).

This ratio comes from the coefficients in the balanced equation. Notice that the ratio was reduced from 8 : 4 to 2 : 1 when placed in the dimensional analysis form. While reducing is not absolutely necessary (the ratio will cancel properly even if not reduced), a good chemistry student notices such things and will do it.

5. Convert moles of SO₂ to grams of SO₂ .

6. All units have been canceled except for grams of SO₂ (product). The problem has been solved. Pick up the calculator and do the math.

Stoichiometry :

The limiting reactant is the reactant that is completely consumed in the reaction.
The limiting reactant is not present in sufficient quantity to react with all other reactants.

The reaction stops when the limiting reactant is completely consumed.

Any remaining reactants are considered “excess reactants”.

The amount of product formed is determined by the “limiting reactant”.
Steps in solving a limiting reactant problem :

Write a balanced equation for the reaction.

Convert both reactant quantities to moles.

Determine the moles of product that could be formed by each reactant.

The least amount in step #3 identifies the limiting reactant.

Use that number of moles of product to determine the mass produced.

A limiting reactant problem example:

What mass of water can be produced by 4 grams of hydrogen gas reacting with 16 grams of oxygen gas?

The problem solution :

1. Write a balanced equation for the reaction.
2 H₂ + O₂ → 2 H₂O

2. Convert both reactant quantities to moles.

3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant.

4. Oxygen produces the least amount of water.
16 grams of Oxygen cannot produce as much water as 4 grams of Hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen.

Oxygen is the “limiting” reactant.

Use oxygen for the calculation of product amount.
5. Complete the problem by converting moles of H₂O to mass of H₂O.

The theoretical yield for this problem is 18 grams. If you performed this reaction in the lab, your actual yield might be less. Can you think of reasons why?

Limiting Reactant Problems
Percent Yield
The quantity of product that is calculated to form when all the limiting reactant is used up is called the theoretical yield.

The amount of product actually obtained in a reaction is called the actual yield.

The actual yield is almost always less than (and never greater than) the theoretical yield

Sample problem :

Given the reaction :
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(aq)

A. If you start with 155 g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe?

B. If the actual yield of Fe was 87.9 g, what was the percent yield?

 

Evaluation :

Balance the following chemical Equations below :

i. S8 + O2 → SO3

ii. HgO → Hg + O

iii. Zn + HCl → H2 + ZnCl2

iv. Na + H2O → NaOH + H2

v. C10H16 + Cl → C + HCl

vi. Si2H3 + O2 → SiO2 + H2O

vii. Fe + O → Fe2O3

viii. FeS2 + O2 → Fe2O3 + SO2

ix. Fe2O3 + H2 → Fe + H2O

x. K + Br →

xi. C2H2 + O2 →

xii. H2O2 → H2O + O2

xiii. C7H16 + O2 → CO2 + H2O

xiv. SiO2 + HF →

xv. KClO3 → KCl + O2

xvi. KClO3 → KClO4 + KCl

xvii. P4O10 + H2O → H3PO4

Sb + O → Sb4O6

Fe2O3 + CO → Fe + CO2

PCl5 + H2O → HCl + H3PO4

H2S + Cl → S8 + HCl

Fe + H2O → Fe3O4 + H2

N + H → NH3

N2 + O2 → N2O

CO2 + H2O → C6H12O6 + O

SiCl4 + H2O → H4SiO4 + HCl

H3PO4 → H4P2O7 + H2O

Al(OH)3 + H2SO4 → Al2(SO4)3 + H2O

Fe2(SO4)3 + KOH → K2SO4 + Fe(OH)3

H2SO4 + HI → H2S + I + H2O

Al + FeO →

P4 + O2 → P2O5

K2O + H2O → KOH

Na2O2 + H2O → NaOH + O

C + H2O → CO + H

H3AsO4 → As2O5 + H2O

Al2(SO4)3 + Ca(OH)2 →

FeCl3 + NH4OH →

Ca3(PO4)2 + SiO2 → P4O10 + CaSiO3

N2O5 + H2O → HNO3

Al + HCl

H3BO3 → H4B6O11 + H2O

Mg + N →

NaOH + Cl → NaCl + NaClO + H2O

Li2O + H2O → LiOH

CaC2 + H2O → C2H2 + Ca(OH)2

Fe(OH)3 → Fe2O3 + H2O

Pb(NO3)2 → PbO + NO2 + O

Ca + AlCl3 → CaCl2 + Al

NH3 + NO → N + H2O

H3PO3 → H3PO4 + PH3

Fe2O3 + C → CO + Fe

FeS + O2 → Fe2O3 + SO2

NH3 + O → NO + H2O

Hg2CO3 → Hg + HgO + CO2

SiC + Cl → SiCl4 + C

Al4C3 + H2O → CH4 + Al(OH)3

Ag2S + KCN → KAg(CN)2 + K2S
Au2S3 + H → Au + H2S

ClO2 + H2O → HClO2 + HClO3

MnO2 + HCl → MnCl2 + H2O + Cl

CONCLUSION : At the end of the lesson learners were able to answer the questions. The Teacher marks their Note books and makes necessary corrections

 

ASSIGNMENT :

1.From XNH3(g)+YO2-ZNO(g)+QH2O(g)
The value of Z is
(A) .4 (B) .7 (C).6 (D).5

2.One molecule of oxygen atoms
(A).has a molar mass of 32g (B).has 6.02x 1023
(C).can be represented as O2 (D) .has a formula mass of 16
(E) Contains Avogadro’s number of atom

3.The numerical coefficients in a balanced equation give
(A).the number of moles of reactants and products
(B).the molar mass of the reactants and products
(C).the number of moles of reactants only
(D).the number of molecules and atoms of products
(E).the mass ratio of the reactants