79.7g of blue vitriol reacts with KI then find the volume of I2 at STP 4CuSO4+4KI=Cu2I2+2K2SO4+I2?

The iodine produced will not be a gas under normal conditions so I will calculate the mass produced.

For a start the equation is not balanced. It should be:

#sf(2CuSO_4+4KIrarr2CuI+2K_2SO_4+I_2)#

Iodine forms in solution with a brown precipitate of copper(I) iodide:

It does not form as a gas at stp.

#sf(M_r[CuSO_4]=159.61)#

#sf(M_r[I_2]=253.81)#

The equations tells us that 2 moles #sf(CuSO_4)# gives 1 mole of #sf(I_2)# molecules.

Converting moles to grams:

#sf(2xx159.61color(white)(x)g" "CuSO_4rarr253.81color(white)(x)g" "I_2)#

#:.##sf(319.22color(white)(x)grarr253.81color(white)(x)g)#

#sf(1color(white)(x)grarr253.81/(319.22)color(white)(x)g)#

#:.##sf(79.7color(white)(x)grarr253.81/(319.22)xx79.7color(white)(x)g)#

#sf(=63.37color(white)(x)g)# of #sf(I_2)#

Another example of a very bad question. If the iodine was produced as a gas at stp (which it isn't) then we would reluctantly have to say:

Moles #sf(I_2=m/M_r=63.37/253.81)#

Assuming the molar volume at stp is 22.4L then :

Volume #sf(I_2=63.37/(253.81)xx22.4=5.6color(white)(x)L)#

This gives option B as the correct answer to an incorrect question.