TEXT BOOK ACTIVITIES - CHEMICAL REACTIONS AND EQUATIONS

Aim: To know the chemical reaction of magnesium burnt in the air.

Materials Required: Magnesium ribbon, Tong, Burner, Watch-glass etc.,

Procedure:

• Clean a magnesium ribbon about 3-4cm long by rubbing it with sandpaper.
• Hold it with a pair of tongs. Burn it using a spirit lamp or burner and collect the ash so formed in a watch glass as shown in the figure, Burn the magnesium ribbon keeping it as far as possible from your eyes.

Observation:Magnesium ribbon burns with a dazzling white light and changes into a white powder. The white powder is magnesium oxide.

Conclusion:

• It is clear that Mg burns and changes into a new substance of a new state, which is magnesium oxide (MgO).
• MgO is formed by the reaction of magnesium with oxygen present in the air.

         Magnesium + oxygen → magnesium oxide

2Mg + O₂ → 2MgO

• The change in state of magnesium ribbon observed in this reaction.

Caution: This activity needs the teacher’s assistance. It would be better if students wear eye protection.

Aim: To know the chemical reaction of potassium iodide and lead nitrate solutions.

Materials Required: Lead nitrate solution, potassium iodide solution, test tubes.

Procedure:

v  Take lead nitrate solution in a test tube. Add potassium iodide solution to this.

v  What do you observe?

Observation:

v  On mixing two solutions a yellow colour precipitate of lead iodide is formed.

         Pb (NO₃)₂ (aq)    + 2KI (aq) → PbI₂ (s)  + 2KNO₃ (s)

 

v  The change in colour  observed in this chemical reaction.

Conclusion: On mixing  Pb(NO₃)₂ and KI, the yellow-coloured precipitate of PbI₂ is obtained.

Aim: To study the reaction between zinc metal and dil. Sulphuric acid.

Materials Required: Zinc granules,  Sulphuric acid, conical flask, test tube, glass tube, cork.

Procedure:

· Take a few zinc granules in a conical flask or a test tube. Add dilute hydrochloric acid or sulphuric acid to this.

· Do you observe anything happening around the zinc granules? Touch the conical flask or test tube. Is there any change in its temperature?

Observation and conclusion:

· Hydrogen gas is produced and the conical flask gets hot due to the action of dil. H₂SO₄. The presence of hydrogen gas can be observed in the form of bubbles accompanied by brisk effervescence.

           Zn(s)     +         H₂SO₄ (aq)       →    ZnSO₄ (aq)      +        H₂(g) ↑

           (Zinc)            (Sulphuric acid)        (Zinc sulphate)           (Hydrogen)

Aim: To understand the reaction between calcium oxide and water.

Materials Required: Calcium oxide and water, beaker.

Procedure:

· Take a small amount of calcium oxide or quick lime in a beaker.

· Slowly add water to this.

· Touch the beaker as shown in figure.  Do you feel any change in temperature?

Observation:

· Calcium oxide and water combine vigorously to produce slaked lime (Calcium hydroxide), releasing a large amount of heat.

·  The reaction is highly vigorous in nature. It is accomplished by a hissing noise, bubbles, and calcium hydroxide (slaked lime) is formed.

· Heat is evolved during the reaction hence; it is an exothermic reaction.

    CaO_(s)            +       H₂O_(l)       →        Ca(OH)_{2 (aq)}

Quick lime                Water                 Slaked lime

Conclusion:  Calcium hydroxide is formed by the combination of calcium oxide and water.

Caution: This activity needs the teacher’s assistance.

Aim: To understand the thermal decomposition reaction.

Materials Required: Ferrous sulphate crystals, boiling tube, burner or spirit lamp.

Procedure:

· Take about 2g of ferrous sulphate crystals in a dry boiling tube.

· Note the colour of the ferrous sulphate crystals.

· Heat the boiling tube over the flame of a burner or spirit lamp as shown in figure below.

· Observe the colour of the crystals after heating.

Observation:

· Ferrous sulphate crystals lose water when heated and the colour of crystals changes.

· Ferrous sulphate crystals are green in colour and have 7 molecules of crystalline water.

                             FeSO₄.7H₂O(s)    →    FeSO₄(s)  +     7H₂O(g)

Conclusion: 

When ferrous sulphate further decomposes, it gives a reddish-brown residue of ferric oxide and fumes of sulphur dioxide and sulphur trioxide evolve.

                         2FeSO_4(s)       →    Fe₂O_{3 (s)}  + SO_{2 (g)}  +SO_{3 (g)}

      The decomposition of ferrous sulphate occurs in the above reaction.

Aim: To understand the thermal decomposition reaction.

Materials Required:  Lead nitrate powder, boiling tube, pair of tongs, burner.

Procedure:

· Take about 2 g of lead nitrate powder in a boiling tube.

· Hold the boiling tube with a pair of tongs and heat it over a flame, as shown in figure.

· what do you observe? Note down the change, if any.

Observation: On heating lead nitrate, reddish-brown fumes with a pungent smell are evolved and a yellow residue remains in the test tube.

Conclusion: On the decomposition of lead nitrate, nitrogen dioxide and oxygen gas are evolved.

2Pb(NO₃)₂ (s)     →      2PbO(s)  +  4NO₂(g)  +   O₂(g)

 Lead nitrate                   Lead oxide 

Aim: To understand the electrolytic decomposition  reaction.

Materials Required:  Plastic mug, water, test tubes, graphite rods, rubber stopper, 6V battery.

Procedure:

· Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in figure.

· Connect these electrodes to a 6-volt battery.

· Fill the mug with water such that the electrodes are immersed. Add a few drops of dilute sulphuric acid to the water. 

· Take two test tubes filled with water and invert them over the two carbon electrodes.

· Switch on the current and leave the apparatus undisturbed for some time.

· You will observe the formation of bubbles at both electrodes. These bubbles displace water in the test tubes. 

· Is the volume of the gas collected the same in both test tubes?

· Once the test tubes are filled with the respective gases, remove them carefully.

· Test these gases one by one by bringing a burning candle close to the mouth of the test tube.

Observation: 

· Bubbles of gases appear above the electrodes in both the test tubes.

· The volume of the gas collected in both the test tubes is not the same. The volume of one of the gases, namely, hydrogen, is twice the volume of the gas(oxygen) collected in the other test tube.

· When we bring a burning candle close to the mouth of one of the test tube, the gas in the test tube catches fire and burns with a pop sound indicating that the gas in the test tube is hydrogen whereas when a burning candle was introduced into the other test tube, the candle started burning brightly indicating that the test tube contained oxygen because the abundance of oxygen accelerates the combustion reaction.

· The gas collected at the anode is oxygen and the gas collected at the cathode is hydrogen.

Conclusion:

· The bubbles produced by the passing current a re hydrogen and oxygen gases which are formed by the decomposition of water on the passing electric current.

· The volume of hydrogen collected will be twice the volume of oxygen. 

· This reaction is called electrolytic decomposition reaction. 

Aim: To understand the photochemical decomposition  reaction.

Materials Required:  Silver chloride, china dish

Procedure:  

· Take about 2 g of silver chloride in a China dish.

· What is its colour?  

· Place this China dish in sunlight for some time.  

· Observe the colour of the silver chloride after some time.

Observation: When silver chloride is placed in the sunlight.  Colour of silver chloride changes.

Conclusion:   

· On the decomposition of silver chloride which is white in colour, it changes to grey.

· Silver chloride decomposes into silver and chlorine. 

              2AgCl (s) + sunlight  ⟶   2Ag(s) + Cl₂(g)

               (White)                                 (grey)

· so, the decomposition reaction takes place in the presence of sunlight, it is called Photolysis.

Aim: To understand the chemical displacement reaction.

Materials Required:  Three iron nails, sand paper, two test tubes, copper sulphate solution.

Procedure:  

· Take three iron nails and clean them by rubbing them with sandpaper.

· Take two test tubes marked (A) and (B). In each test tube, take about 10 ml of copper sulphate solution.

· Tie two iron nails with a thread and immerse them carefully in the copper sulphate solution in test tube B for about 20 minutes, Keep one iron nail aside for comparison.

· After 20 minutes, take out the iron nails from the copper sulphate solution.

· Compare the intensity of the blue colour of copper sulphate solutions in test tubes (A) and (B).

· Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept aside.

Observation:

· After 20 minutes we take out both the nails from test tube B.

· Now we compare both nails with the nail kept aside, and we find the iron nail that remained suspended has a brownish coating on its surface.

· we also find that the blue colour of the copper sulphate solution fades and changes to light green colour which is different from the colour of the copper sulphate solution in test tube A.

       Fe(s)   +     CuSO₄(aq)       →        FeSO₄(aq)      +  Cu(s)

                  (Copper sulphate)          (Iron sulphate)    

·  In the above reaction, iron displaces copper from the copper sulphate solution. 

Conclusion:

This activity shows the displacement of copper from copper sulphate by iron.

Aim: To understand the chemical double displacement reaction.

Materials Required:  sodium sulphate, barium chloride, test tubes.

Procedure:  

· Take about 3 ml of sodium sulphate solution in a test tube.

· In another test tube. take about 3 ml of barium chloride solution.

· Mix the two solutions. (figure)   What do you observe?

Observation:  in this reaction.  it is observed that an insoluble white solid substance is formed.

Conclusion:  

· The insoluble white substance formed is known as a precipitate.

                 Na₂SO₄(aq) + BaCI₂(aq) →     BaSO₄(s) + 2NaCI(aq) 

                 This is also known as Precipitation Reaction.

· The white precipitate of BaSO₄ is formed by  the reaction of SO₄^2- and Ba ²⁺. The other product formed is sodium chloride which remains in the solution.

· Such reaction in which there is an exchange of ions between the reactants are called double displacement reactions. 

Aim: To understand the oxidation and reduction reactions.

Materials Required:  china dish, copper powder.

Procedure:  

· Heat a china dish containing about 1g copper powder, (figure)

· What do you observe?

Observation: 

 On heating, a black coating is formed on the surface of copper.

Conclusion: 

· The surface of copper powder becomes coated with black copper oxide. This is because of the oxidation of copper.

                    2Cu       +      O₂    ⟶           2CuO₂    

                 Copper         Oxygen         Copper(II) oxide

· If we pass hydrogen gas overheated copper oxide (CuO) the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.

· In the above reaction, copper is oxidized to copper oxide, and oxygen is reduced to copper oxide. This is a redox reaction.