The copper(II) ion in a copper(II) sulfate solution reacts with potassium iodide to produce the triiodide ion, I3-. This reaction is commonly used...

The very step in solving this question is to write a well-balanced chemical equation for the given reaction. It can be written as:

`2CuSO_4 + 5KI -> 2CuI + KI_3 + 2K_2SO_4`

Here, 2 moles of copper sulfate reacts with 5 moles of potassium iodide to form 2 moles of copper iodide, 1 mole of potassium triiodide and 2 moles of potassium sulfate. 

Molar mass of each of the species is given as:

CuSO4 = 63.55 + 32.06 + 4 x 16 = 159.6 g/mole

KI = 39.1 + 126.9 = 166 g/mole

CuI = 63.55 + 126.9 = 190.45 g/mole

KI3 = 39.1 + 3 x 126.9 = 419.8 g/mole

K2SO4 = 2 x 39.1 + 32.06 + 4 x 16 = 174.26 g/mole

It is given that 2 gm of KI is added, that is, 0.012 moles (= 2/166) of KI is added. It is added to 0.525 g CuSO4 or 0.0033 moles of CuSO4.

Since, 2 moles of CuSO4 reacts with 5 moles of KI, CuSO4 is the limiting reactant and it will react with only 0.0083 moles of KI (= 5/2 x 0.0033) and this will leave out 0.0037 moles of KI.

The amount of products formed can be calculated by stoichiometry.

CuI :  2 moles CuSO4 produces 2 moles CuI, which means, 0.0033 moles of CuI will be generated, or 0.63 gm (=0.0033 x 190.45) of CuI will be produced.

KI3: 2 moles CuSO4 generates 1 moles of KI3, which means 0.00165 moles of KI3 will be produced, or 0.693 g (= 0.00165 x 419.8) of KI3 will be generated.

K2SO4: 2 moles of CuSO4 generates 2 moles K2SO4, which means 0.0033 moles of K2SO4 will be generated, or 0.575 g (= 0.0033 x 174.26) will be generated.

Hope this helps.