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Answers
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Here is your answer
(a) C12H22O11----->12C + 11H2O
(b) 2NaBr + 2H2SO4
------> Br2 + Na2So4 + So2 +2H2O
(c)2Kcl + H2So4----->2HCl + K2So4
(d) Cu + 2H2So4
------>CuSo4 + So2 + H2o
(e) 3S + 2H2So4------>3So2 + 2H2o
Thanks for question!!
Given H2SO4 is added on 4 test tube having :-
1) cane sugar
2)Sodium bromide
3) Copper turnings
4) Sulphur
Q.1. Formation of black substance.
Ans:- When conc. H2SO4 is react with cane sugar it act as dehydrating agent which removes water from cane sugar and give it black colour this is highly exothermic reaction.
*C12H22O11 (sugar) + H2SO4 (sulfuric acid) → 12 C (carbon) + 11 H2O (water) + mixture of water (black ).
Q.2.Evolution of brown gas?
Ans:- When conc. H2SO4 is added with sodium bromide it oxidised bromine with evolution of brown gas .
Reaction:-
2NaBr+2H2SO4→Na2SO4+Br2+SO2+2H2O .
Q.3.Evolution of colourless gas?
Ans:- When conc.H2SO4 react with sulphur it get oxidised and reduced to form a colourless gas of SO2.
*S (s) + 2H2SO4 (l) → 3SO2 (g) + 2H2O (l)
Q.4.Formation of brown substance which on dillution become blue?
Ans:- When conc. H2SO4 is react with copper turnings it form CuSO4 which is blue in colour.
2H2SO4 + Cu (s) → CuSO4 + SO2 + 2H2O.
Q.5.Disappearance of yellow colour and formation of colourless gas?
Ans:- when conc. H2SO4 react with sulphur the yellow colour of sulphur get faded away with the formation of colourless SO2 gas.
*S (s) + 2H2SO4 (l) → 3SO2 (g) + 2H2O (l)