Question

Suppose of iron(II) chloride is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

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.

      $FeCl_{2} + 2AgNO_{3} \rightarrow Fe(NO_{3})_{2} + 2AgCl(s)$

So, moles of Iron chloride = 5.49 g of molar mass of Iron chloride

              No. of moles = $\frac{mass}{\text{molar mass}}$

                                    = $\frac{5.49 g}{162.2 g/mol}$

                                     = 0.0338 moles $FeCl_{2}$

And, moles of silver nitrate = molarity × volume in L

                                    = 66.0 Mm

or,                                = 0.066 M × 350 ml

                                    = 0.0231 moles $AgNO_{3}$

Therefore, here silver nitrate is the limiting agent .

Now, we will calculate the moles of $Cl^{-}$ which are reacted as follows:.

          $0.0231 moles AgNO_{3} \times \frac{1 moles FeCl_{2}}{2 moles AgNO_{2}}$

                             = 0.01155 moles $FeCl_{2}$

Remaining moles of $FeCl_{2}$ = 0.0338 moles $FeCl_{2}$ - 0.01155 moles $FeCl_{2}$

                        = 0.02225 mole $FeCl_{2}$

Or, 0.0445 moles $Cl^{-}$ ions.

Now, we will calculate the molarity as follows.

                  Molarity = $\frac{\text{number of moles}}{\text{volume in L }}$

                             = $\frac{0.0445 moles Cl^{-}}{0.350 L }$

                             = 0.127 M

Thus, we can conclude that final molarity of chloride anion in the solution is 0.127 M.