Question

Determine the ammount (g) of solid produced when 27.00 mL of 0.2500 M cobalt (III) iodate reacts with 24.00 mL of 0.2700 M ammonium sulfide.

Answer 1

Moles of cobalt(III) iodate = 27.00 mL x 0.2500 M = 6.75 mmol = 6.75 x 10^-3 mol     ( 1 mmol = 10^-3 mol)

Moles of ammonium sulfide = 24.00 mL x 0.2700 M = 6.48 mmol = 6.48 x 10^-3 mol

Balanced equation: 2Co(IO₃)₃(aq) + 3(NH₄)₂S(aq) Co₂S₃(s) + 6NH₄IO₃(aq)

For complete reaction the required mole ratio of Co(IO₃)₃ : (NH₄)₂S = 2 : 3

                                                                                                             = 1 : 1.5

Available mole ratio of Co(IO₃)₃ : (NH₄)₂S = 6.75 : 6.48

                                                                    = 1 : 0.96

Hence, (NH₄)₂S is the limiting reagent in this reaction.

Now, theoretically, 3 mol of (NH₄)₂S gives 1 mol of the solid cobalt(III) sulfide

Therefore, 6.48 x 10^-3 mol of (NH₄)₂S will give (1 mol x 6.48 x 10^-3 mol )/3 mol = 2.16 x 10^-3 mol of the solid cobalt(III) sulfide

Molar mass of cobalt(III) sulfide (Co₂S₃) = 214 g/mol

Hence, the amount of solid Co₂S₃ produced in the reaction = 2.16 x 10^-3 mol x 214 g/mol = 0.462 g