The question is incomplete, the complete question is;
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
Answer:
0.014 M
Step-by-step explanation:
We have to first obtain the number of moles of FeI2
Molar mass of FeI2= 310 g/mol
So n = 0.981/310 = 0.0031 mol
Volume of solution = 150mL or 0.15L
Concentration of AgNO3 = 35mM * 10^-3 = 0.035M
number of moles of AgNO3 = 0.035 x 0.15 = 0.00525 mol
es
Equation of the reaction;
2AgNO3 + FeI2 -------> 2AgI + Fe(NO3
)2
Amount of excess FeI2 = 0.00525 - 0.0031 = 0.00215mol
Concentration of excess iodide in solution = Concentration of excess FeI2/volume of solution = 0.00215/0.15 = 0.0143M