JK-TK/Newdata huggingface dataset:
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. White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087
50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.
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. White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087
50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow. (a)Identify the: (i)cation responsible for the green solution T Cr3+ (ii)possible anions present in white precipitate R CO32-, SO32-, SO42- (b)Name gas V Sulphur (IV)oxide (c)Write a possible ionic equation for the formation of white precipitate R.
Add acidified Barium nitrate(V)/chloride.

White precipitate formed with sodium sulphate (VI) No white precipitate formed with sodium sulphate (IV) (b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3) Chemical equation Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g) Mole ratio Cu(s: SO2 (g) = 1:1 Method 1 1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3 (120 x 1000) g -> (120 x 1000) g x 22.4.dm3) 63.5 g = 42330.7087
50 Method 2 Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles 63.5 Moles SO2 = Moles of Cu = 1889.7639 moles Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4) = 42330.7114 jgthungu@gmail.com162Green solution TWhite precipitate RAcidified K2Cr2O7Barium nitrate(V)Colourless gasVDilute nitric (V) acid4.Use the reaction scheme below to answer the questions that follow.

(a)Identify the: (i)cation responsible for the green solution T Cr3+ (ii)possible anions present in white precipitate R CO32-, SO32-, SO42- (b)Name gas V Sulphur (IV)oxide (c)Write a possible ionic equation for the formation of white precipitate R.

Ba2+ (aq) + CO32- (aq) -> BaCO3(s) Ba2+ (aq) + SO32- (aq) -> BaSO3(s) Ba2+ (aq) + SO42- (aq) -> BaSO4(s)
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