Answer : The mass of sodium iodide used to produced must be, 105.22 grams.
Explanation : Given,
Mass of
= 89.1 g
Molar mass of
= 253.8 g/mole
Molar mass of
= 149.89 g/mole
First we have to calculate the moles of
.
![\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](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DI_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DI_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DI_2%7D%3D%5Cfrac%7B89.1g%7D%7B253.8g%2Fmole%7D%3D0.351moles)
Now we have to calculate the moles of
.
The balanced chemical reaction is,
![2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)](https://tex.z-dn.net/?f=2NaI%28aq%29%2BCl_2%28g%29%5Crightarrow%20I_2%28s%29%2B2NaCl%28aq%29)
From the balanced chemical reaction, we conclude that
As, 1 mole of
obtained from 2 moles of ![NaI](https://tex.z-dn.net/?f=NaI)
So, 0.351 moles of
obtained from
moles of ![NaI](https://tex.z-dn.net/?f=NaI)
Now we have to calculate the mass of
.
![\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DNaI%3D%5Ctext%7BMoles%20of%20%7DNaI%5Ctimes%20%5Ctext%7BMolar%20mass%20of%20%7DNaI)
![\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20%7DNaI%3D%280.702mole%29%5Ctimes%20%28149.89g%2Fmole%29%3D105.22g)
Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.