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Answer to Question #17306 in Inorganic Chemistry for Manolo Javier
Question #17306
Solid Iodine is added to a strongly basic solution of mercury (II) nitrate. What are the products and how do you solve it?
Expert's answer
If ahighly alkaline solution add-crystal iodine, the following reactions:
3I2 + 6KOH = 5KI + KIO3+ 3H2O
Next KI reacts with nitrate of mercury to form a yellow precipitate:
Hg(NO3)2 + 2KI = HgI2 + 2 KNO3
3I2 + 6KOH = 5KI + KIO3+ 3H2O
Next KI reacts with nitrate of mercury to form a yellow precipitate:
Hg(NO3)2 + 2KI = HgI2 + 2 KNO3
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