Answer : The molarity of chloride anion in the solution is 0.003318 mole/L
Explanation : Given,
Mass of
= 0.701 g
Volume of solution = 300 ml = 0.3 L
Molarity of AgCl = 14.0 M = 14.0 M
Molar mass of
= 126.751 g/mole
First we have to calculate the moles of
.
![\text{Moles of }FeCl_2=\frac{\text{Mass of }FeCl_2}{\text{Molar mass of }FeCl_2}=\frac{0.701g}{126.751g/mole}=0.00553moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DFeCl_2%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DFeCl_2%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DFeCl_2%7D%3D%5Cfrac%7B0.701g%7D%7B126.751g%2Fmole%7D%3D0.00553moles)
Now we have to calculate the moles of
.
![\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}=14.0mole/L\times 0.3L=4.2mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DAgNO_3%3D%5Ctext%7BMolarity%20of%20%7DAgNO_3%5Ctimes%20%5Ctext%7BVolume%20of%20solution%7D%3D14.0mole%2FL%5Ctimes%200.3L%3D4.2mole)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
![FeCl_2(aq)+2AgNO_3(aq)\rightarrow 2AgCl(s)+Fe(NO_3)_2(aq)](https://tex.z-dn.net/?f=FeCl_2%28aq%29%2B2AgNO_3%28aq%29%5Crightarrow%202AgCl%28s%29%2BFe%28NO_3%29_2%28aq%29)
From the balanced reaction we conclude that
As, 1 moles of
react with 2 mole of ![AgNO_3](https://tex.z-dn.net/?f=AgNO_3)
So, 0.00553 moles of
react with
moles of ![AgNO_3](https://tex.z-dn.net/?f=AgNO_3)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
.
As, 1 moles of
react to give 2 moles of ![AgCl](https://tex.z-dn.net/?f=AgCl)
So, 0.00553 moles of
react to give
moles of ![AgCl](https://tex.z-dn.net/?f=AgCl)
Now we have to calculate the molarity of
.
![\text{Molarity of }AgCl=\frac{\text{Moles of }AgCl}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20%7DAgCl%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DAgCl%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
![\text{Molarity of }AgCl=\frac{0.01106mole}{0.3L}=0.003318mole/L](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20%7DAgCl%3D%5Cfrac%7B0.01106mole%7D%7B0.3L%7D%3D0.003318mole%2FL)
As we know that, 1 mole of AgCl in solution gives 1 mole of silver ion and 1 mole of chloride ion.
So, the molarity of chloride ion = Molarity of AgCl = 0.003318 mole/L
Therefore, the molarity of chloride anion in the solution is 0.003318 mole/L