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What happens when Cl_(2) is passed through a hot concentrated solution of a base like Ba(OH)_(2). |
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Answer» Solution :On treatment with `Ba(OH)_(2), Cl_(2)` undergoes disproportionation to form barium choride, `BaCl_(2)` and barium chlorate, `Ba(ClO_(3))_(2)`. `6 Ba(OH)_(2)(conc.)+6 overset(0)(Cl_(2))overset("Hot") RARR 5 overset(-1)(BaCl_(2))+overset(+5)(Ba(ClO_(3))_(2))+6 H_(2)O` |
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