Iodine is prepared both in the laboratory and commercially by adding Cl 2 (g) Cl2 (g) to an aqueous solution containing sodium iodide. 2 NaI (aq) + Cl 2 (g) ⟶ I 2 (s) + 2 NaCl (aq) 2NaI (aq) + Cl2 (g) ⟶I2 (s) + 2NaCl (aq) How many grams of sodium iodide, NaI, NaI, must be used to produce 83.9 g 83.9 g of iodine, I 2?

The reaction of Chlorine with Sodium Iodide is given as;

2NaI + Cl2 - I2 + 2NaCl

The molar mass of iodine gas is 253.8089g/mol.

The molar mass of Sodium Iodide is 149.89g/mol.

The number of moles of I2 can be calculated as shown below;

moles of I2 = Given mass (g) / molecular mass (g/mol).

= 83.9g/253.8089g/mol.

= 0.330mol

In the above reaction, 2mol of NaI reacts with 1 mol of Cl to form 1 mol of I2 and 2 moles of NaCl.

The number of moles of NaI would be;

Number of moles of NaI = moles of I2 * 2 mol of NaI/1 mol of I2

= 0.330mol * 2/1

= 0.660mol

Therefore, the mass of NaI needed would be;

Mass of NaI = moles of NaI * molar mass of NaI

= 0.660mol * 149.89g/mol

= 98.92g

The mass of Sodium Iodide required is 98.92g.