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

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.