SALTMODULE 2

Gas Test Result of test

 Ammonia  Place a damp

in the gas

 Red litmus paper turns blue

 Carbon dioxide

 Bubble the gas through   

Lime water turns   

TESTING FOR GASES

red litmus paper

limewater milky

 Chlorine  Place a piece of damp

 in the gas 

 Blue litmus paper turns red, then is  

 Hydrogen

 Put a   wooden splint near the gas 

Gas burns with

blue litmus paper

bleached

lighted

‘pop’ sound

 Oxygen  Put a  wooden splint near the gas

 Glowing splinter is  

 Hydrogen chloride

 Dip a glass rod into concentrated ammonia, NH3 solution, bring a drop of ammonia to the mouth of the test tube. 

 Dense

are observed

glowingignited

white fumes

Sulphur dioxide

Bubble the gas through a

solution OrBubble the gas through a

solution

  colour decolorized 

Acidified potassium manganate(VII)

Acidified potassium dichromate(VI)

Purple

Orange solutionchange to green solution

Glowingwooden splinter Burning

wooden splinter

Limewater

to blue

red

colourless

potassium manganate(VII)

con.ammonia

red

Action of heat on carbonate salts1. All carbonates decompose on heating except ……………………………………  

1.Most carbonate decompose on heating to produce ………………….and ……… 

Ammonium, sodium and potassium carbonate

Metal oxide Carbon dioxide

 1.Write a chemical equation for reaction of zinc carbonate

when heated

 1.How can you identify the gas that is produced when a

carbonate metal is heated? ………………………………………………………………

23 COZnOZnCO

Bubble the gas through limewater, it will turns cloudy

  Colour of carbonate salt 

Colour of oxide formed

A. zinc carbonate

   

 

B . lead(II) carbonate

   

 

C . copper(II) carbonate

   

 

whiteYellow whenHot,white whencold

white

green

brown whenhot,yellow whencold

black

  Colour of carbonate salt 

Colour of oxide formed

D .calcium carbonate

   

 

E . magnesium carbonate

   

 

F . Aluminium carbonate

   

 white

white

white

white

white

white

ACTION OF HEAT ON NITRATE SALTS

1) All nitrate decompose on heating to produce metal oxide , nitrogen dioxide and oxygen gas

Eg. Mg(NO3)2 MgO + NO2 + O2

Mg(NO3)2 MgO + 2NO2 + ½ O2

2Mg(NO3)2 2MgO + 4NO2 + O2

1) All nitrate decompose on heating to produce metal oxide , nitrogen dioxide and oxygen gas

2Cu(NO3)2 2CuO + 4NO2 + O2

2Zn(NO3)2 2ZnO + 4NO2 + O2

2Pb(NO3)2 2PbO + 4NO2 + O2

How to identify the gas produced when nitrate salt is heated?

a) OXYGEN GAS , O2: INSERT / PUT GLOWING WOODEN SPLINTER at the mouth of TEST TUBE containing OXYGEN GAS.

AND it will IGNITE/ BURN

Method, substance apparatus

result

b) NITROGEN DIOXIDE GAS, NO2 : PUT / PLACE DAMP BLUE LITMUS PAPER on the mouth of TEST TUBE containing NITROGEN DIOXIDE GAS. AND it will TURN RED

How to identify the gas produced when nitrate salt is heated?

Method, substance apparatus

result

Colour changes

COLOUR OF NITRATE SALT

COLOUR OF OXIDE FORMED

A.Copper(II) nitrate

B. Zinc nitrate

BLUE SALT CRYSTAL

BLACK SOLID

WHITE SALT CRYSTAL

YELLOW WHEN HOTWHITE WHEN COLD

SOLVE THESE PROBLEMS

SALT X SOLID Y GAS W THAT TURNS LIMEWATER MILKY

SALT X DISSOLVES FORMING COLOURLESS SOLUTION P

WHITE PRECIPITATE THAT DISSOLVES IN EXCESS AMMONIA SOLUTION

HEAT

ADD DILUTE HNO3

ADD NH3 SOLUTION

1) NAME GAS W

2) STATE THE COLOUR OF SOLID Y

3) IDENTIFY SALT X

4) Write chemical equation of salt X when heated

CARBON DIOXIDE

YELLOW WHEN HOT WHITE WHEN COLD

ZnCO3 / zinc carbonate

ZnCO3 ZnO + CO2

If 12.5 g of salt X was heated to produce solid Y and gas W, calculate :[ RAM X is 125 and Y is 81. Molar volume is 24dm3mol-1 ]

i) THE MASS OF SOLID Y FORMED

Mol ZnCO3 = mass/ molar mass = 12.5/ 125 = 0.1 moleFrom equation, 1 mol ZnCO3 produced 1 Mol of ZnO.

Therefore , 0.1 mol ZnCO3 produced 0.1 mol ZnOMass ZnO = 0.1 mol x 81 = 8.1 g

ii) THE volume of gas W FORMED

MOL ZnCO3 = mass/ molar mass = 12.5/ 125 = 0.1 mole

From equation, 1 mol ZnCO3 produced 1 Mol of CO2.So , 0.1 mol ZnCO3 produced 0.1 mol CO2

Mass ZnO = 0.1 mol x 24 = 2.4 dm3

Salt R1consist of one cation and one anion.The chemical test is done to identified the cation and anion present in R1. Based on procedure and observation given, state the inference in the space provided

TEST PROCEDURE OBSERV INFERENCEWATER, H2O

Pour distilled water into test tube . Add R1 salt and shake well

R1 do not dissolve in water

R1 is insoluble salt

Salt R1consist of one cation and one anion.The chemical test is done to identified the cation and anion present in R1. Based on procedure and observation given, state the inference in the space provided

TEST PROCEDURE OBSERV INFERENCENitric acid, HNO3

Pour nitric acid into test tube.Add R1Pass the gas through lime water

Bubbles of gas release

Limewater turn chalky

CO2 is released.Carbonate ion is present in R1

Test Procedure Observation Inference

NaOH Add a few drop of NaOH to R1, shake.Add NaOH until excess and shake well

White precipitate formed

Dissolves in excess

Pb2+ or Zn2+ or Al3+

might present

Test Procedure Observation Inference

NH3 Add a few drop of NH3 solution to R1, shake.Add NH3 solution until excess and shake well

White precipitate formedWhite precipitate do not dissolves in excess NH3

solution

Pb2+ or Al3+

might present

Test Procedure Observation Inference

KI Add KI to R1 and shake well

Yellow precipitate is formed

Present Pb2+

SALT R1 CONTAIN :

CATION : Pb2+

ANION : CO32-

volume

various

Draw a graph volume of potassium iodide against height of precipitate formed from the above table

b) What is the colour of the precipitate formed? …………………………………………………………………………………………….

c) Name the precipitate formed ……………………………………………………………………………..

yellow

Lead (II) iodide

Calculate the number of moles of lead (II) nitrate solution in 5 cm 3

No of mole = MV/1000 = 0.5 x 5 / 1000

= 0.0025 mol

What is the volume of potassium iodide that exactly react with 5.00 cm 3 of lead (II) nitrate solution?

(volume that exactly react is the volume when the height of precipitate start constant)

Volume = 5.00 cm3

Calculate the number of moles of potassium iodide that reacts with 5 cm3 lead (II) nitrate solution

No of mole = MV/1000 = 1.0 x 5 / 1000

= 0.005 mol

Deduce empirical formula of the precipitate formed

mole Pb2+ : mole I-

0.0025 : 0.0050

1:2

PbI2

Hence, construct ionic equation for the reaction above

Pb2+ + 2I- PbI2