Suppose 1.90g of iron(II) iodide is dissolved in 350.mL of a 64.0mM 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) iodide is dissolved in it.

Round your answer to 3 significant digits.

0.0697 M

Explanation:

From the balanced equation:

1 mole of FeI2 requires 2 moles of AgNO3

mole of AgNO3 = molarity x volume

             = 0.064 x 0.350 = 0.0244 mole

Mole of FeI2 required = 0.0244 x 1/2

                                      = 0.0112 mole

Mole of FeI2 present = mass/molar mass

                                   = 1.97/309.65

                                      = 0.006362

Hence, FeI2 is the limiting reagent.

2 moles of AgNO3 is gives 1 mole of Fe(NO3)2

0.0244 mole of AgNO3 gives = 0.0233 x 1/2 = 0.0112 mole of Fe(NO3)2

There are 2 moles of Iron (II) cation for every 1 mole of Fe(NO3)2

mole of iron (II) cation in 0.0112 mole Fe(NO3)2:

                       = 2 x 0.0112 = 0.0244 mole

Molarity of iron (II) cation = mole/volume

                                          = 0.0244/0.350 = 0.06971

To 3 significant figure = 0.0697 M