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SALT. MODULE 2. TESTING FOR GASES. red litmus paper. limewater. milky. blue litmus paper. bleached. lighted. ‘pop’ sound. glowing. ignited. white fumes. Acidified potassium manganate(VII). Purple. Orange solution change to green solution. Acidified potassium dichromate(VI).
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SALT MODULE 2
TESTING FOR GASES red litmus paper limewater milky
blue litmus paper bleached lighted ‘pop’ sound
glowing ignited white fumes
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 • How can you identify the gas that is produced when a carbonate metal is heated? • ……………………………………………………………… Bubble the gas through limewater, it will turns cloudy
Yellow when Hot,white when cold white brown when hot,yellow when cold white green black
white white white white white white
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? • 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 WHITE SALT CRYSTAL YELLOW WHEN HOT WHITE WHEN COLD
SOLVE THESE PROBLEMS HEAT SOLID Y SALT X GAS W THAT TURNS LIMEWATER MILKY ADD DILUTE HNO3 ADD NH3 SOLUTION SALT X DISSOLVES FORMING COLOURLESS SOLUTION P WHITE PRECIPITATE THAT DISSOLVES IN EXCESS AMMONIA SOLUTION
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 ] • 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
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 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 CO2 is released. Carbonate ion is present in R1
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 b) What is the colour of the precipitate formed? ……………………………………………………………………………………………. c) Name the precipitate formed …………………………………………………………………………….. 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