Respuesta :
Answer : The molarity of chloride anion in the solution is 0.003318 mole/L
Explanation : Given,
Mass of [tex]FeCl_2[/tex] = 0.701 g
Volume of solution = 300 ml = 0.3 L
Molarity of AgCl = 14.0 M = 14.0 M
Molar mass of [tex]FeCl_2[/tex] = 126.751 g/mole
First we have to calculate the moles of [tex]FeCl_2[/tex].
[tex]\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[/tex]
Now we have to calculate the moles of [tex]AgNO_3[/tex].
[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}=14.0mole/L\times 0.3L=4.2mole[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]FeCl_2(aq)+2AgNO_3(aq)\rightarrow 2AgCl(s)+Fe(NO_3)_2(aq)[/tex]
From the balanced reaction we conclude that
As, 1 moles of [tex]FeCl_2[/tex] react with 2 mole of [tex]AgNO_3[/tex]
So, 0.00553 moles of [tex]FeCl_2[/tex] react with [tex]0.00553\times 2=0.01106[/tex] moles of [tex]AgNO_3[/tex]
From this we conclude that, [tex]AgNO_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]FeCl_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AgCl[/tex].
As, 1 moles of [tex]FeCl_2[/tex] react to give 2 moles of [tex]AgCl[/tex]
So, 0.00553 moles of [tex]FeCl_2[/tex] react to give [tex]0.00553\times 2=0.01106[/tex] moles of [tex]AgCl[/tex]
Now we have to calculate the molarity of [tex]AgCl[/tex].
[tex]\text{Molarity of }AgCl=\frac{\text{Moles of }AgCl}{\text{Volume of solution}}[/tex]
[tex]\text{Molarity of }AgCl=\frac{0.01106mole}{0.3L}=0.003318mole/L[/tex]
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