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 67.3 g of iodine, I2?

79.0 g

Explanation:

1. Gather the information in one place.

MM: 148.89 253.81

2NaI + Cl2 → I2 + 2NaCl

m/g: 67.3

2. Moles of I2

n = 67.3 g * (1 mol/253.81 g) = 0.2652 mol I2

2. Moles of NaI needed

From the balanced equation, the molar ratio is 2 mol NaI: 1 mol I2

n = 0.028 76 mol I2 * (2 mol NaI/1 mol I2) = 0.5303 mol NaI

3. Mass of NaI

m = 0.5303 mol * (148.89 g/1 mol) = 79.0 g NaI

It takes 79.0 g of NaI to produce 67.3 g of I2.