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Concentrated sulphuric acid is added followed by heating in each of the following test tubes labelled (i) to (v) : Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. (a) formation of black substance (b) evolution of brown gas (c ) evolution of colourless gas (d) formation of brown subtance which on dilution becomes blue. (e ) disappearance of yellow powder along with the evolution of a colourless gas. |
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Answer» SOLUTION :(a) Black substance will be formed in tube (i), Actually, concentrated sulphuric ACID is a powerful dehydrating agent. It dehydrates sugar `(C_(12)H_(22)O_(11))` completely to leave a black residue of carbon with the smell of burning sugar. This is called CHARRING of sugar. `underset(("Sugar"))(C_(12)H_(22)O_(11)) overset("Conc."H_(2)SO_(4)) underset("heat") to underset(("Black mass")) (12C+ 11H_(2)O)` (b) Brown gas will evolve in tube (ii). It is bromine formed on heating sodium bromide with concentrated sulphuric acid. The gas is actually yellowish brown in colour. `2NaBr+3H_(2)SO_(4) overset("heat") to2NaHSO_(4)+SO_(2)+Br_(2)+2H_(2)O` (c ) Colourless gas will evolve in tube (v). It is hydrogen CHLORIDE gas formed by heating POTASSIUM chloride with concentrated sulphuric acid. `KCl+H_(2)SO_(4) overset("heat") to KHSO_(4)+HCl(g)` (d) Brown substance cupric oxide (CuO) will be formed in the tube (iii) It will further react with conc. `H_(2)SO_(4)` to form copper sulphate `(CuSO_(4))` to give a blue solution. `Cu+H_(2)SO_(4) to CuO +SO_(2)+H_(2)O` `CuO+H_(2)SO_(4) to CuSO_(4)+H_(2)O` `ul(Cu+2H_(2)SO_(4) to CuSO_(4)+SO_(2)+2H_(2)O)` (e ) Yellow powder of sulphur will disappearin tube (iv) on reacting with concentrated sulphuric acid. Sulphur dioxide, a colourless gas with pungent smell will evolve during the reaction `H_(2)SO_(4)to H_(2)O+SO_(2)+(O)xx16` `(S_(8)16O to 8SO_(2))/(S_(8)+16 H_(2)SO_(4) to 24SO_(2)+16H_(2)O)` |
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