Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution, You can assume the volume of the solution doesn't change shen th s sove m Be sure your answer has the correct number of significant digits

Final molarity of iodide ion C (I-) = 0.0143M

Explanation:

n = (m (FeI (2))) / (M (FeI (2))

Molar mass of FeI (3) = 55.85 + (127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V (solution) = 150mL = 0.15L

C (AgNO3) = 35mM = 0.035M = 0.035m/L

n (AgNO3) = C (AgNO3) x V (solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI (3) = AgI (3) + FeNO3

So, n (FeI (3)) excess = 0.00525 - 0.0031 = 0.00215mol

C (I-) = C (FeI (3)) = [n (FeI (3)) excess] / [V (solution) ] = 0.00215/0.15 = 0.0143mol/L or 0.0143M