SALT MODULE 2
TESTING FOR GASES 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 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 glowing ignited white fumes
Sulphur dioxide Bubble the gas through a solution Or Bubble the gas through a solution colour decolorized Acidified potassium manganate(VII) Purple Orange solution change to green solution Acidified potassium dichromate(VI)
Glowing wooden splinter Burning wooden splinter to blue Limewater
red con.ammonia colourless red potassium manganate(VII)
Action of heat on carbonate salts 1. All carbonates decompose on heating except …………………………………… Most carbonate decompose on heating to produce ………………….and ……… Ammonium, sodium and potassium carbonate Metal oxide Carbon dioxide
Write a chemical equation for reaction of zinc carbonate when heated Write a chemical equation for reaction of zinc carbonate when heated How can you identify the gas that is produced when a carbonate metal is heated? ……………………………………………………………… Bubble the gas through limewater, it will turns cloudy
white white green black Yellow when Hot,white when cold brown when Colour of carbonate salt Colour of oxide formed A. zinc carbonate B . lead(II) C . copper(II) Yellow when Hot,white when cold white brown when hot,yellow when cold white green black
white white white white white white Colour of carbonate salt Colour of carbonate salt Colour of oxide formed D .calcium carbonate E . magnesium F . Aluminium 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
2Cu(NO3)2 2CuO + 4NO2 + O2 2Zn(NO3)2 2ZnO + 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? 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
How to identify the gas produced when nitrate salt is heated? 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 Method, substance apparatus result
Colour changes BLUE SALT CRYSTAL BLACK SOLID COLOUR OF NITRATE SALT COLOUR OF OXIDE FORMED Copper(II) nitrate B. Zinc nitrate BLUE SALT CRYSTAL BLACK SOLID WHITE SALT CRYSTAL YELLOW WHEN HOT WHITE WHEN COLD
SOLVE THESE PROBLEMS HEAT 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
CARBON DIOXIDE YELLOW WHEN HOT WHITE WHEN COLD ZnCO3 / zinc carbonate 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
Mol ZnCO3 = mass/ molar mass = 12.5/ 125 = 0.1 mole 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 ] THE MASS OF SOLID Y FORMED Mol ZnCO3 = mass/ molar mass = 12.5/ 125 = 0.1 mole From equation, 1 mol ZnCO3 produced 1 Mol of ZnO. Therefore , 0.1 mol ZnCO3 produced 0.1 mol ZnO Mass 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
TEST PROCEDURE OBSERV INFERENCE 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 INFERENCE WATER, H2O Pour distilled water into test tube . Add R1 salt and shake well R1 do not dissolve in water R1 is insoluble salt
TEST PROCEDURE OBSERV INFERENCE 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 INFERENCE Nitric acid, HNO3 Pour nitric acid into test tube. Add R1 Pass 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 formed White 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
yellow Lead (II) iodide c) Name the precipitate formed 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 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