Question 1
Give a chemical test to distinguish between the following gases:
(a) Hydrogen and oxygen
(b) Carbon dioxide and sulphur dioxide
(c) Hydrogen chloride and hydrogen sulphide
(d) Chlorine and nitrogen dioxide
(e) Ammonia and hydrogen chloride
(f) Sulphur dioxide and chlorine
(a) Hydrogen and oxygen
Burning wooden splinter is extinguised in hydrogen whereas Oxygen rekindles a glowing wooden splinter.
(b) Carbon dioxide and sulphur dioxide
Carbon dioxide has no effect on acidified KMnO4 or K2Cr2O7.
Sulphur dioxide turns KMnO4 from pink to clear colourless and K2Cr2O7 from orange to clear green.
2KMnO4 + 2H2O + 5SO2 ⟶ K2SO4 + 2MnSO4 + 2H2SO4
K2Cr2O7 + H2SO4 + 3SO2 ⟶ K2SO4 + Cr2(SO4)3 + H2O
(c) Hydrogen chloride and hydrogen sulphide
Hydrogen chloride — Forms a curdy white precipitate on passage through AgNO3 solution.
AgNO3 [aq.] + HCl ⟶ AgCl ↓ [curdy white ppt.] + HNO3
Hydrogen sulphide — Turns moist lead acetate paper silvery black.
Pb(CH3COO)2 [colourless] + H2S ⟶ PbS ↓ [black] + 2CH3COOH
(d) Chlorine and nitrogen dioxide
Chlorine reacts with silver nitrate solution to form a white precipitate of silver chloride.
4AgNO3 [aq.] + 2Cl2 (g) ⟶ 4AgCl ↓ [curdy white ppt.] + 2N2O5 (g) + O2 (g)
Nitrogen dioxide does not react with silver nitrate solution, hence, no white precipitate is formed.
(e) Ammonia and hydrogen chloride
Ammonia turns Nessler's reagent from colourless to pale brown.
Hydrogen chloride shows no action with Nessler's reagent. It forms a curdy white ppt. on passage through AgNO3 solution.
AgNO3 [ag.] + HCl ⟶ AgCl ↓ [curdy white ppt.] + HNO3
The ppt. of AgCl is soluble in NH4OH but insoluble in dil. HNO3
(f) Sulphur dioxide and chlorine
On passing Sulphur dioxide gas through lime water, it turns lime water milky.
Ca(OH)2 + SO2 ⟶ CaSO3 ↓ [white ppt.] + H2O
Chlorine gas does not turn lime water milky. It turns moist starch iodide paper blue black.
Cl2 + 2KI ⟶ 2KCl + I2
Starch + I2 ⟶ Blue black colour