Question

Suppose 1.35g of iron(II) chloride is dissolved in 250.mL of a 29.0mM 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.Round your answer to 3 significant digits.

Answer 1

Number of moles of FeCl2 = 1.35/126.75 = 0.01065 moles

number of moles of AgNO3 = 29 m(moles/L) * 250 mL = 7.25 mmoles

FeCl2 + 2 AgNO3 -------> 2 AgCl + Fe(NO3)2

2 moles of AgNO3 require 1 moles of FeCl2

7.25 mmoles AgNo3 requires 7.25/2 * 10^-3moles of FeCl2

                         = 0.003625 moles

Number of moles remaining = 0.01065 - 0.003625 = 0.007025 moles

Concentration of Cl- ion or FeCl2 solution = 0.007025*1000/250

Concentration of Cl- ions or FeCl2 solution = 0.0281 mol/L