Question

A reaction is carried out by combining 65.00 mL of a 0.108 M sample of (NH4)3PO4 and 48.00 mL of a 0.230 M Fe9NO3)2 solution. The net ionic equation is :

3Fe^2+(aq) + 2PO4^3-(aq) ---> Fe3(PO4)2(s) balanced

1a) What is the mass of iron(II) phosphate (MM= 357.49 g/mol) produced in the reaction?

1b) What are the concentrations of the reacting ions (Fe^2+(aq) and PO4^3+(aq)) at the end of the reaction (assuming no change in volume)?

Answer 1

3 Fe(NO₃)₂ + 2 (NH₄)₃PO₄ = Fe₃(PO₄)₂(s) + 6 (NH₄)NO_{3 ................i}

Complete Ionic Reaction is written as

3Fe²⁺ + 6NO₃^- + 6NH₄+ + 2PO₄^3- = Fe₃(PO₄)₂(s) + 6NO₃^- + 6NH₄⁺...................ii

And Net Ionic Equation is

3Fe²⁺(aq) + 2PO₄^3-(aq) ---> Fe₃(PO₄)₂(s).................iii

Now given that

[(NH₄)₃PO₄] = 0.108 and V= 65 ml

Moles of (NH₄)₃PO₄ = 0.108 mol/L*0.065 L = 0.00702 mol

Fe(NO₃)₂ =0.230 M , V= 48 mol

Moles of Fe(NO3)2 = 0.230 * 0.048 =0.011 mol

3Fe²⁺ + 6NO₃^- + 6NH₄⁺ + 2PO₄^3- = Fe₃(PO₄)₂(s) + 6NO₃^- + 6NH₄⁺

Now complete ionic reaction shows that

3 Fe(NO₃)₂ + 2 (NH₄)₃PO₄ = Fe₃(PO₄)₂(s) + 6 (NH₄)NO₃

From reaction 2: 3 mol of Fe(NO₃)₂ gives 2 mol Fe²⁺ and 6 mol NO3-

So for 0.011 mol Fe(NO₃)₂ , Fe²⁺ = 0.0073 mol and NO₃- = 0.022 mol

From reaction 2: 2 mol of (NH₄)₃PO_4, gives 6 mol NH₄⁺ and 2 mol PO₄^3-

So, for 0.00702 mol of (NH₄)₃PO₄ 0.0210 mol NH₄⁺ and 0.00702 mol PO₄^3-

Now

Net reaction

3Fe^2+(aq) + 2PO4^3-(aq) ---> Fe3(PO4)2(s)

From this reaction: 3 mol Fe2+ will react with 2 mol PO43- to form 1 mol Fe3(PO4)2

We calcuated that initial concentrations of Fe2+ = 0.0073 mol and PO43- = 0.00702 mol

Here limiting reactant is Fe2+

So, 0.073 mol of Fe2+ will react with 0.0049 mol of PO₄^3-to form 0.00243 mol Fe3(PO4)2

From reaction 2:

a) Total 0.00243 mol Fe3(PO4)2 will form in this reaction.

Mass of Fe3(PO4)2 produced = 0.00243 mol*357.49 g/mol = 0.869 g

b) Final concentrations of Fe2+(aq) & PO₄^3- are

Fe2+ is limiting so, Fe2+ concentration= 0

Final moles of PO₄^3- = Initial-consumed = 0.00702-0.0049 = 0.00212 mol

Total volume is 48+65 m= 113 ml= 0.113 L

Final concentration of Fe2+= 0

Final concentration of PO43- = 0.00212/0.113 = 0.0188 M