We are adding iron (II) chloride to a solution of silver nitrate . The reaction that occurs between them is
The complete ionic equation that takes place can be written as
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 =
Total volume of the solution, V = 250. mL = 0.250 L
Hence concentration of FeCl2 after dissolving is
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)
please answer question correctly and show your work! Suppose 2.26 g of Iron(II) chloride is dissolved...
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