Suppose of iron(II) iodide is dissolved in of a 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 when the iron(II) iodide is dissolved in it. Be sure your answer has the correct number of significant digits.

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

0.014 M

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

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it means there are 3 atoms of oxygen