Question

Kmno4 reacts with KI in basic medium to form di-iodine(I2) and MnO2 and when 250 ml of 0.1 molar KI solutions are mixed with 250 ml of 0.02 molar kMnO4 in basic medium what is the number of moles of di-iodine(I2) formed

Answer 1

Given  250 ml of 0.1M  K I   =>  0.025 moles 

and    250 ml  of 0.02M  of KMnO₄  =>  0.005 moles

I am not sure of the reaction that takes place.... We need to find the products and then balance equation first.

    2 K Mn O₄ + 6 K I   => 2 Mn O₂  +  3 I₂   + 4 K₂ O

 So   n moles of Potassium Permanganate reacts with 3 n moles of potassium iodide to produce 1.5 n moles of Iodine gas.

So 0.005 moles of permanganate will react with 0.015 moles of Iodide, and produce 0.0075 moles of Iodine gas.

If the reaction is 

   2 K MnO₄ + 6 K I + 4 H₂ O   ==>   3 I₂ + 2 Mn O₂  + 6 K OH

then,

      n moles of permanganate will react with 3 n moles of Iodide and produce 1.5 moles of Iodine....  Hence we will have  0.0075 moles of Iodine gas.

Answer 2

Answer:

Explanation:

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