Question

Suppose 0.938g of iron(II) chloride is dissolved in 300.mL of a 46.0mM aqueous solution of silver nitrate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) chloride is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Balanced chemical reaction;

FeCl2(aq) + 2AgNO3(aq) -----> 2AgCl(s) + Fe(NO3)2(aq)

Moles of FeCl2 = mass/molar mass = 0.938/126.751 = 0.0074 moles

Moles of AgNO3 = Molarity * volume in liters = 46*10^-3 * (300/1000) = 0.0138 moles

From reaction;

2 Mole AgNO3 requires 1 mole FeCl2

So, 0.0138 mole AgNO3 will require = 0.0138/2 = 0.0069 moles of FeCl2

Remaining mole of FeCl2 in solution = 0.0074 - 0.0069 = 0.005 moles

FeCl2 ----> Fe2+ + 2Cl-

1 mole FeCl2 contains 2 moles Cl-

So, 0.0005 mole will contain = 2 * 0.0005 = 0.001 mole of Cl-

Volume = 300 ml = 0.300 liters

Molarity of Cl- = 0.001/0.300 = 0.00333 M ...Answer

Let me know if any doubts.