If in the given reaction, 3I 2 + OH- ---> IO3 - + 5I-
2 moles of iodine are taken, then the ratio of iodate and iodide ions formed in the alkaline medium is
Why doesn't it matters?
It doesn't matter how much was reacted, the ratio is the same, as given in the balanced equation. 1:5
Can anybody tell me the detailed answer?
To find the ratio of iodate (IO3-) to iodide (I-) ions formed in the given reaction, you need to calculate the stoichiometric coefficients of iodate and iodide ions.
The balanced chemical equation for the reaction is:
3I2 + 6OH- ---> 2IO3- + 10I-
Based on the equation, the molar ratio between iodine (I2) and iodate (IO3-) ions is 3:2, and the molar ratio between iodine (I2) and iodide (I-) ions is 5:10 or 1:2.
Since you have started with 2 moles of iodine (I2), you can calculate the amount of iodate (IO3-) and iodide (I-) ions formed.
For iodate (IO3-) ions:
2 moles I2 x (2 moles IO3- / 3 moles I2) = 4/3 moles IO3-
For iodide (I-) ions:
2 moles I2 x (10 moles I- / 5 moles I2) = 4 moles I-
Therefore, the ratio of iodate (IO3-) to iodide (I-) ions formed in the alkaline medium is 4/3:4, which simplifies to 1:3.