JK-TK/Newdata huggingface dataset:
Old newspapers turn brown on exposure to air on regaining the atomic oxygen. The bleaching through adding oxygen by chlorine gas is permanent. (b)Turns Orange acidified potassium dichromate(VI) to green Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or; (ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing Sulphur(IV) oxide gas. Observation: Orange acidified potassium dichromate(VI) turns to green. Explanation: Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq) 3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l) This is a confirmatory test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue. (c)Decolorizes acidified potassium manganate(VII)
15 Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas.
Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue. (c)Decolorizes acidified potassium manganate(VII)
15 Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or; (ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas. Observation: Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized. Explanation: Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO4- ions to green Mn2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of MnO4- (aq) 5SO32-(aq) + 2MnO4- (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l) (purple) (colourless) This is another test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO4- ions to colourless Mn2+ ions leaving a yellow residue. (d)Decolorizes bromine water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.
(d)Decolorizes bromine water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or; (ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl. Observation: Yellow bromine water turns to colourless/ bromine water is decolorized. Explanation:
16 Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue itself oxidized from SO32- ions in sulphuric (IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of MnO4- (aq) SO32-(aq) + Br2 (aq) + H2O(l) -> SO42-(aq) + 2HBr(aq) (yellow) (colourless) This can also be used as another test for the presence of Sulphur(IV) oxide gas. Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue. (e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green Explanation: Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
(e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+ Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or; (ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl. Observation: Yellow/brown Iron (III)chloride solution turns to green Explanation: Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: (i)Reaction of Sulphur(IV) oxide gas with water SO2(g) + H2O(l) -> H2 SO3 (aq) (ii)Dissociation /ionization of Sulphuric(IV)acid. H2 SO3 (aq) -> 2H+(aq) + SO32- (aq) (iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq) SO32-(aq) + 2Fe3+ (aq) +3H2O(l) -> SO42-(aq) + 2Fe2+(aq) + 2H+(aq) (yellow) (green)
17 (f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas Experiment: (i)Pass a stream of Sulphur(IV) oxide 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 Sulphur(IV) oxide gas. Swirl. Observation: Brown fumes of a gas evolved/produced. Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Swirl. Observation: Brown fumes of a gas evolved/produced. Explanation: Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. Chemical/ionic equation: SO2(g) + 2HNO3 (l) -> H2 SO4 (l) + NO2 (g) (brown fumes/gas) (g)Reduces Hydrogen peroxide to water Experiment: (i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or Barium chloride followed by five drops of 2M hydrochloric acid/ 2M nitric(V) acid. Observation: A white precipitate is formed that persist /remains on adding 2M hydrochloric acid/ 2M nitric(V) acid. Explanation: Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid. When Ba2+ ions in Barium Nitrate(V) or Barium chloride solution is added, a white precipitate of insoluble Barium salts is formed showing the presence of of either SO32- ,SO42- ,CO32- ions. i.e. Chemical/ionic equation: SO32-(aq) + Ba2+ (aq) -> BaSO3(s) white precipitate SO42-(aq) + Ba2+ (aq) -> BaSO4(s) white precipitate CO32-(aq) + Ba2+ (aq) -> BaCO3(s) white precipitate
18 If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate: (i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present. (ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because: I.
Chemical/ionic equation: SO32-(aq) + Ba2+ (aq) -> BaSO3(s) white precipitate SO42-(aq) + Ba2+ (aq) -> BaSO4(s) white precipitate CO32-(aq) + Ba2+ (aq) -> BaCO3(s) white precipitate
18 If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate: (i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present. (ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because: I. BaSO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO2 gas that turns Orange moist filter paper dipped in acidified Potassium dichromate to green. Chemical equation BaSO3(s) +2H+(aq) -> Ba2+ (aq) + SO2(g) + H2O(l) I. BaCO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic CO2 gas that forms a white precipitate when bubbled in lime water. Chemical equation BaCO3(s) +2H+(aq) -> Ba2+ (aq) + CO2(g) + H2O(l) 5.Sulphur(IV)oxide also act as an oxidizing agent as in the following examples. (a)Reduction by burning Magnesium Experiment Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas Observation Magnesium ribbon continues to burn with difficulty. White ash and yellow powder/speck Explanation Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen. The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.
Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen. The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash. Yellow specks of sulphur residue form on the sides of reaction flask/gas jar. During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the metal is oxidized (reducing agent) Chemical equation SO2(g) + 2Mg(s) -> 2MgO(s) + S(s) (white ash/solid) (yellow speck/powder)
19 (b)Reduction by Hydrogen sulphide gas Experiment Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas. Or Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl Observation Yellow powder/speck Explanation Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar. A little moisture/water act as catalyst /speeds up the reaction. Chemical equation SO2(g) + 2H2S(g) -> 2H2O(l) + 3S(s) (yellow speck/powder) 6.Sulphur(IV)oxide has many industrial uses. State three. (i)In the contact process for the manufacture of Sulphuric(VI)acid (ii)As a bleaching agent of pulp and paper. (iii)As a fungicide to kill microbes’ (iv)As a preservative of jam, juices to prevent fermentation (ii) Sulphur(VI)oxide(SO3) (a) Occurrence Sulphur (VI)oxide is does not occur free in nature/atmosphere (b) Preparation In a Chemistry school laboratory Sulphur (VI)oxide may prepared from: Method 1;Catalytic oxidation of sulphur(IV)oxide gas. Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid .
(i)In the contact process for the manufacture of Sulphuric(VI)acid (ii)As a bleaching agent of pulp and paper. (iii)As a fungicide to kill microbes’ (iv)As a preservative of jam, juices to prevent fermentation (ii) Sulphur(VI)oxide(SO3) (a) Occurrence Sulphur (VI)oxide is does not occur free in nature/atmosphere (b) Preparation In a Chemistry school laboratory Sulphur (VI)oxide may prepared from: Method 1;Catalytic oxidation of sulphur(IV)oxide gas. Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid . The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form Sulphur (VI)oxide gas. Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a freezing mixture /ice cold water. 20 The solid fumes out on heating to a highly acidic poisonous gas. Chemical equation 2SO2(g) + O2(g) --platinised asbestos--> 2SO3 (g) Method 2; Heating Iron(II)sulphate(VI) heptahydrate When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a boiling tube ,it loses the water of crystallization and colour changes from green to white. Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) (green solid) (white solid) On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas. Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III)oxide is left as a brown residue/solid.
Chemical equation FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l) (green solid) (white solid) On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas. Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water. Iron(III)oxide is left as a brown residue/solid. Chemical equation 2FeSO4 (s) -> Fe2O3(s) + SO2 (g) + SO3(g) (green solid) (brown solid) Caution On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI)acid and thus its preparation in a school laboratory is very risky. (c) Uses of sulphur(VI)oxide One of the main uses of sulphur(VI)oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale/production of sulphuric(VI)acid. (iii) Sulphuric(VI)acid(H2SO4) (a) Occurrence Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three mineral acids; Nitric(V)acid Sulphuric(VI)acid Hydrochloric acid. 21 Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level. (b)The Contact process for industrial manufacture of H2SO4 . I. Raw materials The main raw materials for industrial preparation of Sulphuric(VI)acid include: (i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS) (ii)Oxygen from fractional distillation of air (iii)Water from rivers/lakes II. Chemical processes The contact process involves four main chemical processes: (i)Production of Sulphur (IV)oxide As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources; I. Burning/roasting sulphur in air.
Raw materials The main raw materials for industrial preparation of Sulphuric(VI)acid include: (i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS) (ii)Oxygen from fractional distillation of air (iii)Water from rivers/lakes II. Chemical processes The contact process involves four main chemical processes: (i)Production of Sulphur (IV)oxide As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources; I. Burning/roasting sulphur in air. Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners Chemical equation S(s) + O2(g) --> SO2 (g) II. Burning/roasting sulphide ores in air. Sulphur (IV)oxide gas is produced as a by product in extraction of some metals like: - Lead from Lead(II)sulphide/Galena,(PbS) - Zinc from zinc(II)sulphide/Zinc blende, (ZnS) - Copper from Copper iron sulphide/Copper pyrites, (CuFeS2) On roasting/burning, large amount /quantity of sulphur(IV)oxide is generated/produced. Chemical equation (i)2PbS (s) + 3O2 (g) -> 2PbO(s) + 2SO2 (g) (ii)2ZnS (s) + 3O2 (g) -> 2ZnO(s) + 2SO2 (g) (ii)2CuFeS2 (s) + 4O2 (g) -> 2FeO(s) + 3SO2 (g) + Cu2O(s) Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely. 22 (ii)Purification of Sulphur(IV)oxide Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface. The impurities are removed by electrostatic precipitation method .
22 (ii)Purification of Sulphur(IV)oxide Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface. The impurities are removed by electrostatic precipitation method . In the contact process Platinum or Vanadium(V)oxide may be used. Vanadium(V)oxide is preferred because it is : (i) cheaper/less expensive (ii) less easily poisoned by impurities (iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC in a heat exchanger. The heated mixture is passed through long pipes coated with pellets of Vanadium (V)oxide catalyst. The close “contact” between the reacting gases and catalyst give the process its name. Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas. Chemical equation 2SO2 (g) + O2(g) -- V2O5 --> 2SO2 (g) This reaction is exothermic (-∆H) and the temperatures need to be maintained at around 450oC to ensure that: (i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450oC (ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450oC. (iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic. To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower.
(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic. To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower. The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid. Chemical equation H2SO4 (aq) + SO3 (g) -> H2S2O7 (l)
23 Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully with distilled water to give concentrated sulphuric (VI) acid . Chemical equation H2S2O7 (l) + H2O (l) -> 2H2SO4 (l) The acid is stored ready for market/sale. III. Environmental effects of contact process Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere. In the Contact process, about 2% of these gases do not form sulphuric (VI) acid. The following precautions prevent/minimize pollution from Contact process: (i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers. (ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water. This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid. (iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3 Chemical equation Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g) IV.
This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid. (iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid. This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3 Chemical equation Ca(OH)2 (aq) + SO2(g) --> CaSO3 (aq) + H2O (g) IV. Uses of Sulphuric(VI)acid Sulphuric (VI) acid is used: (i) in making dyes and paint (ii)as acid in Lead-acid accumulator/battery (iii) for making soapless detergents (iv) for making sulphate agricultural fertilizers VI. Sketch chart diagram showing the Contact process
24 (c) Properties of Concentrated sulphuric(VI)acid (i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC. (ii) Concentrated sulphuric(VI)acid is very soluble in water. The solubility /dissolution of the acid very highly exothermic. The concentrated acid should thus be diluted slowly in excess water. Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container. (iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+ ions. Free H+ ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change the blue litmus paper red. (iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution. This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases. (v)The following are some chemical properties of concentrated sulphuric (VI) acid: I. As a dehydrating agent Experiment I; Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker. Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.
(v)The following are some chemical properties of concentrated sulphuric (VI) acid: I. As a dehydrating agent Experiment I; Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker. Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes. Observation; Colour( in brown sugar )change from brown to black. Colour (in glucose) change from white to black. 10cm3 beaker becomes very hot. Explanation
25 Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically and physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds. When added to sugar /glucose a vigorous reaction that is highly exothermic take place. The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon. Caution This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi. Chemical equation Glucose: C6H12O6(s) --conc.H2SO4--> 6C (s) + 6H2O(l) (white) (black) Sugar: C12H22O11(s) --conc.H2SO4--> 12C (s) +11H2O(l) (brown) (black) Experiment II; Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Warm . Observation; Colour change from blue to white. Explanation Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds. The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate . Chemical equation CuSO4.5H2O(s) --conc.H2SO4--> CuSO4 (s) + 5H2O(l) (blue) (white) Experiment III; Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid.