********QUESTION******Percent Yield . Write a complete and balanced equation for

Question

********QUESTION******Percent Yield. Write a complete and balanced equation for the synthesis of the coordination compound. Use iron(II) oxalate (the yellow/orange solid) as a reactant. In other words,

FeC2O4(s) + H2O2(aq) + ??? product(s)

Assume that the net reaction occurs in an acidic solution: this will make balancing the equation easier. Use the real formula of the iron (III) oxalate compound that you found in the literature to balance the equation and determine the theoretical yield. You should write out the individual half reactions. (need help with this***) Calculate the percent yield of the hydrated potassium iron(III) oxalate: remember that Fe(NH4)2(SO4)2•6H2O is the limiting reactant.

Background:

The coordiantion compound is synthesized from a reaction mixture of iron(II) ammonium sulfate hexahydrate, Fe(NH4)2(SO4)2•6H2O, oxalic acid, H2C2O4, potassium oxalate, K2C2O4, and hydrogen peroxide, H2O2. In the first step, oxalate ion from oxalic acid precipitates the iron(II) ion as FeC2O4.

Fe(NH4)2(SO4)2·6H2O (aq) + H2C2O4 (aq) + + 2- (1) FeC2O4 (s) + 2NH4 (aq) +2H (aq) +2SO4 (aq) +6H2O(l)

In the second step, iron(II) is oxidized to iron(III) with the formation of the desired product by 1st treating the FeC2O4 precipitate with acid, and then adding an excess of oxalate ions, as potassium oxalate, for bonding to the iron(III) ion in the complex. Hydrogen peroxide, a mild oxidizing agent, oxidizes all iron(II) ions to iron(III) ions. The soluble hydrated potassium iron(III) oxalate compound forms light-sensitive emerald green crystals when a less polar solvent such as alcohol is added to the solution.

(2)

FeC2O4 (s) + K+(aq) + C2O42-(aq) + H2O2(aq) KwFex(C2O4)y(H2O)z (s) + ??????

This reaction is shown incomplete and unbalanced

Explanation / Answer

To synthesize K3[Fe(C2O4)3]·3H2O from Ferrous ammonium sulfate, Fe(NH4)2(SO4)2·6H2O, by dissolving in a slightly acid solution, excess oxalic acid, H2C2O4, is added, and the following reaction takes place:

Fe(NH4)2(SO4)2· 6H2O + H2C2O4 = FeC2O4(s) + H2SO4 + (NH4)2SO4 + 6H2O(l)

(Eq 1)

Potassium oxalate is added to the FeC2O4 precipitate which produces a slightly basic solution for the oxidation of the ferrous ion to the ferric ion by hydrogen peroxide, H2O2. The following reaction takes place:

2Fe 2+

=

H2O + HO2- + 2e-

=

H2O + HO2- + 2Fe 2+

=

(Eq 2)

The OH- ion concentration of the solution is high enough so that some of the Fe3+ reacts with OH- to form ferric hydroxide (brown precipitate) as follows:

Fe3+ + 3OH- = Fe(OH)3(s)

(Eq 3)

With the addition of more H2C2O4, the Fe(OH)3 dissolves and the soluble complex K3[Fe(C2O4)3]· 3H2O is formed according to:

3K2C2O4 + 2Fe(OH)3(s) +3H2C2O4 = 2K3[Fe(C2O4)3]· 3H2O + 3H2O

To calculate % yield of hydrated pottasium iron oxalate:

(Eq 4)

K2C2O4

Fe(NH4)2(SO4)2·6H2O

K3[Fe(C2O4)3]·3H2O

1 mole of Ferrous ammonium sulfate = 1 mole of Fe2+

392.1 g of ferrous ammonium sulfate = 55.8 g Fe

2 moles of Fe2+ = 2 moles of Fe3+ = 2 moles of potassium iron(3) complex

2*55.8 g Fe = 2*491.1 g potassium iron 3 complex = 982.2 g of complex

Theoritical yield of K3[Fe(C2O4)3]·3H2O formed is 982.2 g starting from 2 moles of Ferrous ammonium salt as a reactant.

Fe(NH4)2(SO4)2· 6H2O + H2C2O4 = FeC2O4(s) + H2SO4 + (NH4)2SO4 + 6H2O(l)

(Eq 1)

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.