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.

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