Question

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Suppose 2.26 g of Iron(II) chloride is dissolved in 250. ml of a 38.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iron(II) cation 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. M

Answer 1

We are adding iron (II) chloride to a solution of silver nitrate . The reaction that occurs between them is

$FeCl_2_{(aq)} + 2AgNO_3_{(aq)} \rightarrow 2AgCl_{(s)}+Fe(NO_3)_2 _{(aq)}$

The complete ionic equation that takes place can be written as

$Cl^-_{(aq)} + Ag^+_{(aq)} \rightarrow AgCl_{(s)}$

Hence, are the spectator ions.

Hence, finally we get a solid precipitate in the form of AgCl.

To calculate the number of Fe(II) cation, since Fe2+ is a spectator ion, its initial concentration after dissolving and before reaction will be same as its final concentration after reaction.

Amount of FeCl2 added = 2.26 g

Molar mass of FeCl2 = 126.751 g/mol

Hence, number of moles of FeCl2 added =

$\frac{mass}{molar \ mass} = \frac{2.26 \ g}{126.751 \ g/mol} \approx 0.01783 \ mol$

Total volume of the solution, V = 250. mL = 0.250 L

Hence concentration of FeCl2 after dissolving is

$\frac{n_{FeCl_2}}{V} = \frac{0.01783 \ mol}{0.250 \ L} \approx 0.0713 \ mol/L ={\color{Red} 0.0713 \ M}$

Note that each mole of FeCl2 produces 1 mol of Fe(II) cation.

Hence, concentration of Fe(II) cation in the solution is 0.0713 M. Since it is a spectator ion in the reaction that occurs, the final molarity of Fe(II) cation in the solution is 0.0713 M. (rounded to 3 significant figures)