Question

Iodine is prepared both in the laboratory and commercially by adding Cl2(g)Cl2(g) to an aqueous solution containing sodium iodide. 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq) How many grams of sodium iodide, NaI,NaI, must be used to produce 89.1 g89.1 g of iodine, I2?I2? mass: g NaI

Answer 1

Answer : The mass of sodium iodide used to produced must be, 105.22 grams.

Explanation : Given,

Mass of $I_2$ = 89.1 g

Molar mass of $I_2$ = 253.8 g/mole

Molar mass of $NaI$ = 149.89 g/mole

First we have to calculate the moles of $I_2$.

$\text{Moles of }I_2=\frac{\text{Mass of }I_2}{\text{Molar mass of }I_2}=\frac{89.1g}{253.8g/mole}=0.351moles$

Now we have to calculate the moles of $NaI$.

The balanced chemical reaction is,

$2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)$

From the balanced chemical reaction, we conclude that

As, 1 mole of $I_2$ obtained from 2 moles of $NaI$

So, 0.351 moles of $I_2$ obtained from $2\times 0.351=0.702$ moles of $NaI$

Now we have to calculate the mass of $NaI$.

$\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI$

$\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g$

Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.