Iodine is prepared both in the laboratory and commercially by adding Cl 2 ( g ) to an aqueous solution containing sodium iodide. 2 NaI ( aq ) + Cl 2 ( g ) ⟶ I 2 ( s ) + 2 NaCl ( aq ) How many grams of sodium iodide, NaI , must be used to produce 86.1 g of iodine, I 2 ?

101.69 grams.

Explanation:

The reaction involved in the preparation of Iodine is mentioned in the question as -

⇒

Looking at the stoichiometric coefficients we can say that 2 moles of is required for the preparation of one mole of . In terms of mass we can say that, 299.78 grams of is required for the production of 253.81 grams of .

Now, since 253.81 grams of is produced by 299.78 grams of

So,

Grams of required for the production of 86.1 grams of =

grams

i think its 5049.

i hope this !