Question

Suppose 1.06g of iron(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

the balanced equation is

FeBr₂ + 2AgNO3 ==> 2AgBr(s) + Fe(NO3)₂

the no. of moles of FeBr₂ = grams/molar mass

                                     = 1.06/215.65      (molar mass of FeBr₂ = 215.65)

                                     = 0.0049 mol or   4.9153 mM

milimols 2AgNO3 = mM x L = 52 x 0.200 = 10.4

for AgNO3 = 10.4/2 = 5.2

since the no. of mM of iron(II) bromide is less than mM of AgNO3.i.e. 4.9153<5.2

so there is no unreacted iron(II) bromide .