Respuesta :
Explanation:
Let us assume that 5.49 g of Iron chloride is dissolved in 350 ml of 66.0 mM aqueous solution of silver nitrate
Hence, the balanced chemical equation for this reaction is as follows.
[tex]FeCl_{2} + 2AgNO_{3} \rightarrow Fe(NO_{3})_{2} + 2AgCl(s)[/tex]
So, moles of Iron chloride = 5.49 g of molar mass of Iron chloride
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{5.49 g}{162.2 g/mol}[/tex]
= 0.0338 moles [tex]FeCl_{2}[/tex]
And, moles of silver nitrate = molarity × volume in L
= 66.0 Mm
or, = 0.066 M × 350 ml
= 0.0231 moles [tex]AgNO_{3}[/tex]
Therefore, here silver nitrate is the limiting agent .
Now, we will calculate the moles of [tex]Cl^{-}[/tex] which are reacted as follows:.
[tex]0.0231 moles AgNO_{3} \times \frac{1 moles FeCl_{2}}{2 moles AgNO_{2}}[/tex]
= 0.01155 moles [tex]FeCl_{2}[/tex]
Remaining moles of [tex]FeCl_{2}[/tex] = 0.0338 moles [tex]FeCl_{2}[/tex] - 0.01155 moles [tex]FeCl_{2}[/tex]
= 0.02225 mole [tex]FeCl_{2}[/tex]
Or, 0.0445 moles [tex]Cl^{-}[/tex] ions.
Now, we will calculate the molarity as follows.
Molarity = [tex]\frac{\text{number of moles}}{\text{volume in L }}[/tex]
= [tex]\frac{0.0445 moles Cl^{-}}{0.350 L }[/tex]
= 0.127 M
Thus, we can conclude that final molarity of chloride anion in the solution is 0.127 M.
