JK-TK/Newdata huggingface dataset:
Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green. Yellow solid solid Explanation: Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur. Chemical/ionic equation: H2S(aq) + 2Fe3+ (aq) -> S (s) + Fe2+(aq) + 2H+(aq) (yellow solution) (yellow residue) (green) (e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or; (ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Brown fumes of a gas evolved/produced. Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized to yellow sulphur. Chemical/ionic equation: H2S(g) + 2HNO3 (l) -> 2H2O(l) + S (s) + 2NO2 (g)
38 (yellow residue) (brown fumes) (f)Reduces sulphuric(VI)acid to Sulphur Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or; (ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar containing Hydrogen sulphide gas. Swirl. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur. Chemical/ionic equation: 3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s) (yellow residue) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide.
Swirl. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur. Chemical/ionic equation: 3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s) (yellow residue) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Observation: Yellow solid residue Explanation: Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and itself oxidized to yellow sulphur Chemical/ionic equation: H2S(g) + H2O2 (l) -> 2H2O(l) + S (s) (yellow residue) 8.Name the salt formed when: (i)equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution: Sodium hydrogen sulphide Chemical/ionic equation: H2S(g) + NaOH (l) -> H2O(l) + NaHS (aq) (ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution: Sodium sulphide Chemical/ionic equation:
39 H2S(g) + 2NaOH (l) -> 2H2O(l) + Na2S (aq) Practice Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V)salts as in the flow chart below (a)Name the black solid Copper(II)sulphide (b)Identify the cation responsible for the formation of: I. Blue solution Cu2+(aq) II. Green solution Fe2+(aq) III. Brown solution Fe3+(aq) (c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide -Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution. -Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green.
Green solution Fe2+(aq) III. Brown solution Fe3+(aq) (c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide -Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution. -Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green. -Yellow solid residue/deposit is formed with Hydrogen sulphide Chemical/ionic equation: 4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (d)State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator. ObservationGas continues burning with a blue flame Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water. Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g) Hydrogen sulphide Blue solution Brown solution Black solid Green solution
40 (v)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts 1. Sulphate (VI) (SO42-) salts are normal and acid salts derived from Sulphuric (VI)acid H2SO4. 2. Sulphate(IV) (SO32-) salts are normal and acid salts derived from Sulphuric (IV)acid H2SO3. 3. Sulphuric (VI)acid H2SO4 is formed when sulphur(VI)oxide gas is bubbled in water. The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4-) salts. i.e.
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO4 (aq) -> 2H+(aq) + SO42-(aq) H2SO4 (aq) -> H+(aq) + HSO4 -(aq) All Sulphate (VI) (SO42-) salts dissolve in water/are soluble except Calcium (II) sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II) sulphate (VI) (PbSO4) All Hydrogen sulphate (VI) (HSO3-) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate (VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as solids. Other Hydrogen sulphate (VI) (HSO4-) salts do not exist except those of Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II) hydrogen sulphate (VI) (Mg (HSO4)2). 4. Sulphuric (IV)acid H2SO3 is formed when sulphur(IV)oxide gas is bubbled in water. The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO3 (aq) -> 2H+(aq) + SO32-(aq)
41 H2SO3 (aq) -> H+(aq) + HSO3 -(aq) All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3) All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water.
It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4-) salts. i.e. H2SO3 (aq) -> 2H+(aq) + SO32-(aq)
41 H2SO3 (aq) -> H+(aq) + HSO3 -(aq) All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3) All Hydrogen sulphate (IV) (HSO3-) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate (IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as solids. Other Hydrogen sulphate (IV) (HSO3-) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO3)2). 5.The following experiments show the effect of heat on sulphate(VI) (SO42-)and sulphate(IV) (SO32-) salts: Experiment: In a clean dry test tube place separately about 1.0g of : Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers. Observations: -Colourless droplets of liquid forms on the cooler parts of the test tube in all cases. -White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV).
Test any gases produced using litmus papers. Observations: -Colourless droplets of liquid forms on the cooler parts of the test tube in all cases. -White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV). -Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI) -Colour changes from blue to white then black in case of Copper (II) sulphate (VI) -Blue litmus paper remain and blue and red litmus paper remain red in case of Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV) -Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI). Explanation (i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g. K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l) Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l) MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l) CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l) ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l) FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
42 CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l) All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI).
Explanation (i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g. K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l) Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l) MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l) CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l) ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l) FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
42 CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l) All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI). (i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a brown /yellow residue. Chemical equation 2FeSO4 (s) -> Fe2O3(s) + SO2(g) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water. Sulphur (IV) oxide does not. (ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas. Chemical equation 2CuSO4 (s) -> CuO(s) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. 6.
(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas. Chemical equation 2CuSO4 (s) -> CuO(s) + SO3(g) This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas. 6. The following experiments show the test for the presence of sulphate (VI) (SO42-)and sulphate(IV) (SO32-) ions in a sample of a salt/compound: Experiments/Observations: (a)Using Lead(II)nitrate(V) I. To about 5cm3 of a salt solution in a test tube add four drops of Lead(II)nitrate(V)solution. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- , Cl- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , Cl- ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.(a)To the preserved sample observation 1 in (II) above, Heat to boil. Observation 1
43 Observation Inference White precipitate/ppt persists on boiling SO42- ions Observation 2 Observation Inference White precipitate/ppt dissolves on boiling Cl - ions .(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I.
Observation 1 Observation Inference White precipitate/ppt persists SO42- , Cl- ions Observation 2 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.(a)To the preserved sample observation 1 in (II) above, Heat to boil. Observation 1
43 Observation Inference White precipitate/ppt persists on boiling SO42- ions Observation 2 Observation Inference White precipitate/ppt dissolves on boiling Cl - ions .(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Experiments/Observations: (b)Using Barium(II)nitrate(V)/ Barium(II)chloride I. To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve. Observation Inference White precipitate/ppt SO42- , SO32- , CO32- ions II. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2
44 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).
To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve. Observation 1 Observation Inference White precipitate/ppt persists SO42- , ions Observation 2
44 Observation Inference White precipitate/ppt dissolves SO32- , CO32- , ions III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI). Observation 1 Observation Inference (i)acidified potassium manganate(VII)decolorized (ii)Orange colour of acidified potassium dichromate(VI) turns to green SO32- ions Observation 2 Observation Inference (i)acidified potassium manganate(VII) not decolorized (ii)Orange colour of acidified potassium dichromate(VI) does not turns to green CO32- ions Explanations Using Lead(II)nitrate(V) (i)Lead(II)nitrate(V) solution reacts with chlorides(Cl-), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV). Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. 45 - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .
Chemical/ionic equation: Pb2+(aq) + Cl- (aq) -> PbCl2(s) Pb2+(aq) + SO42+ (aq) -> PbSO4 (s) Pb2+(aq) + SO32+ (aq) -> PbSO3 (s) Pb2+(aq) + CO32+ (aq) -> PbCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists. 45 - Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g) PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g) (iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed; - Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling) - Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling. (iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.
Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. Using Barium(II)nitrate(V)/ Barium(II)Chloride (i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV). 46 Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. .
46 Chemical/ionic equation: Ba2+(aq) + SO42+ (aq) -> BaSO4 (s) Ba2+(aq) + SO32+ (aq) -> BaSO3 (s) Ba2+(aq) + CO32+ (aq) -> BaCO3 (s) (ii)When the insoluble precipitates are acidified with nitric(V) acid, - Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists. - Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively. . Chemical/ionic equation: BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + SO2 (g) BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + CO2 (g) (iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced; - sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.
Carbon(IV)oxide will not. Chemical equation: 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) - Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1.
Sulphur(IV)oxide will not. Chemical equation: Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l) These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water. 47 Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts Practice revision question 1. Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl
48 (a)Identify the: I: Sodium salt solution Sodium sulphate(IV)/Na2SO3 II: White precipitate Barium sulphate(IV)/BaSO3 III: Gas G Sulphur (IV)Oxide /SO2 IV: Colourless solution H Barium chloride /BaCl2 (b)Write an ionic equation for the formation of: I.White precipitate Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s) II.Gas G Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq) III. Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow.
Study the flow chart below and use it to answer the questions that follow Unknown salt Lead(II)nitrate(V) White precipitates of Cl-, SO42- , SO32- and CO32- Dilute nitric(V) acid white ppt dissolves in SO32- and CO32- white ppt persist /remains in SO32- and CO32- Acidified KMnO4 decolorized in SO32- White ppt with lime water in CO32- Acidified KMnO4 K2Cr2O7 / Lime water White ppt dissolves on heating in Cl- White ppt persist on heating in SO42- 2Heat to boil Sodium salt solution White precipitate Barium nitrate(VI) Colourless solution B Gas G and colour of solution changes orange to green Acidified K2Cr2O7 Dilute HCl
48 (a)Identify the: I: Sodium salt solution Sodium sulphate(IV)/Na2SO3 II: White precipitate Barium sulphate(IV)/BaSO3 III: Gas G Sulphur (IV)Oxide /SO2 IV: Colourless solution H Barium chloride /BaCl2 (b)Write an ionic equation for the formation of: I.White precipitate Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s) II.Gas G Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq) III. Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow. 49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I.
Green solution from the orange solution 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) (Orange) (green) 2. Study the flow chart below and answer the questions that follow. 49 (i)Write equation for the reaction taking place at: I.The roasting furnace (1mk) 2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g) II.The absorption tower (1mk) H2SO4 (l) + SO3 (g) -> H2S2O7(l) III.The diluter (1mk) H2S2O7(l) + H2 O(l) -> 2H2SO4 (l) (ii)The reaction taking place in chamber K is SO2 (g) + 1/2O2 (g) SO3 (g) I. Explain why it is necessary to use excess air in chamber K To ensure all the SO2 reacts II.Name another substance used in chamber K Vanadium(V)oxide 3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI). Add acidified Barium nitrate(V)/chloride.