Question

Iodine is prepared both in the laboratory and commercially by adding Cl2(g) to an aqueous solution containing sodium iodide.

2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq)

How many grams of sodium iodide, NaI, must be used to produce 78.7 g of iodine, I2?

Answer 1

Using mole ratios
2NaI=I2(s)

For every 2 moles of sodium iodide you will have one mole of Iodine.

Change grams iodine to moles iodine

n=m/M

n=number of moles
m=mass in grams,
M=molar mass.

n=78.7/254
n=0.309…

Now use mole ratios
0.309..I2x 2NaI/I2 (another way of writing ratio)

The iodine cancels out.

0.618 moles NaI

Then change back to grams.
using n=m/M
0.618=m/150

m=92.7g NaI

m=92.7 NaI is needed to produce 78.7g I2

Molar mass I2
127x2 =254 g/mol I3


Molar mass NaI
MNaI=23+127
MNaI=150 g/mol.