Question

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 ?

Answer 1

Answer:

101.69 grams.

Explanation:

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

$2 NaI_(aq_) +Cl_2_(g_)$ ⇒ $I_2_(s_) +2NaCl_(aq_)$

Looking at the stoichiometric coefficients we can say that 2 moles of $NaI$ is required for the preparation of one mole of $I_2$. In terms of mass we can say that, 299.78 grams of $NaI$ is required for the production of 253.81 grams of $I_2$.

Now, since 253.81 grams of $I_2$ is produced by 299.78 grams of $NaI$

So,

Grams of $NaI$ required for the production of 86.1 grams of $I_2$ =

$\frac{299.78\times86.1}{253.81} =101.69$ grams