CHEMICAL EQUATION

 

Week:   ONE                         Date:                                                   Time:

Period:                                   Duration:                                           Average age of learners:

Subject:                               CHEMISTRY                                                         Class:  SS ONE

Topic:                                                    CHEMICAL EQUATION

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA[ 4th Edition]  pg49-57

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials: beam balance

Entry behavior: The students have been taught linear equation in mathematics

Behavioural objective: At the end of the lesson the students should be able to:

1st PERIOD

Explain chemical equation

Write balance chemical equation

2nd  PERIOD

Balance chemical equation

3rd PERIOD

Solve chemical equation problem

CHEMICAL EQUATION

1ST PERIOD

STEP I: The teacher explains chemical equation

Chemical equations are representations of chemical reactions in term of symbols and formulae of the elements and compounds involved.

A chemical equation is said to be balanced when there are the same number of the same type of every atom on both sides of the equation.

Getting a balanced equation

Balanced symbol equations show what happens to the different atoms in reactions. For example, copper and oxygen react together to make copper oxide.

Take a look at this word equation for the reaction:

Copper and oxygen are the reactants because they are on the left of the arrow. Copper oxide is the product because it is on the right of the arrow.

If we just replace the words shown above by the correct chemical formulae, we will get an unbalanced equation, as shown here:

Notice that there are unequal numbers of each type of atom on the left-hand side compared with the right-hand side. To make things equal, you need to adjust the number of units of some of the substances until you get equal numbers of each type of atom on both sides.

Here is the balanced symbol equation:

 Balancing equations

To balance equations on your own, follow these simple rules:

Check that all the formulae in the equation are correct.

Deal with only one element at a time.

Balancing is adding BIG numbers. You cannot change any of the small numbers in a chemical formula. If balancing is required, put the number in front of the substance.

Check each element again and repeat step 3 again if needed.

Example

This equation is unbalanced. There are four carbon atoms on the left hand side and only one on the right. To balance the carbon, add a big '4' in front of the carbon dioxide.

Next, to balance the hydrogen. We have 8 atoms of hydrogen on the left hand side in C 4H8and only 2 on the right hand side. To balance the hydrogen atoms, add a big '4' in front of H2O.

We’re not finished yet. Now that the carbon and hydrogen have been balanced, we only have to balance the oxygen. We have 2 atoms of oxygen on the left, but in total on the right (taking into account what we have balanced already) we have 12 oxygen atoms. This can be balanced by adding a big '6' in front of the diatomic oxygen molecule on the left hand side.

The balanced equation will be:

Evaluation: The teacher evaluates the lesson by asking the following questions-

Explain chemical equation

State the steps involved in balancing chemical equation

2nd PERIOD

Step I – The teachers leads the students to  balance some chemical equation

NaOH + HCl  → NaCl + H2O

KClO3   → KCl + O2

NaOH  + CO2 →  Na2CO3 +  H2O

Evaluation: The teacher evaluates the lesson by giving these exercise to the students

Balance the following equations

C  + H2O   →  CO + H2

BaCl2  + H2SO4 →  BaSO4 + HCl

NH3 + O2  → H2O  + N2

Al(OH)3  + HNO3  →  Al(NO3)3  +  H2O

H2SO4  +  KOH  →  K2SO4  +  H2O

NH3  +  O2  →  NO  +  H2O

ZnCO3  +  HCl  →  ZnCl  + H2O + CO2

3rd  PERIOD

The teacher leads the students in solving problem from chemical equations

Question

2NaHCO3(s)  → Na2CO3(s) + CO2(g) + H2O(g).

From the equation above, calculate the mass of the residue formed if 1.68g of NaHCO3 is decomposed. [Na= 23, C= 12, O=16, H= 1] .

Solution

Relative  molecular masses of the relevant reactant and product

2NaHCO3    →                             Na2CO3        +                       CO2                         + H2O.

                                      2(23+1+12+48)                              (46+12+48)                         (12+32)             (2+16)

                                                168g                                             106   g                                       44g                       18g

                                                168g of  NaHCO3  produced 106g of Na2CO3(s)

                                                                        1.68g  of  NaHCO3  produce     of Na2CO3(s)

                                                                                                                = 1.06g

Consider the reaction represented by the equation.

Mg(s) + 2HCl(aq)                   MgCl2(aq) + H2

What mass of hydrogen would be produced if  12.0g of magnesium react with excess dilute hydrochloric acid.

[ H= 1, Mg = 24, Cl = 35.5]

Solution

Mg(s)                                  +               2HCl(aq)                               MgCl2(aq)               +                 H2

                                24g                                         2(1+35.5)g                           [24+(35.5)2]g                     ( 1x2)g

                                24g of Mg produced  2g   of    H2

                                12g  of  Mg produce    of  H2 

                                                                                                                        = 1.0g of H2

 Assignment

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