In: Chemistry
Why would the actual coefficients of a ferrous ammonium sulfate hexahydrate and the potassium iron oxalate complex in a balanced chemical equation describing the overall process not be required to compute the predicted weight of the product?
Ans: Ferrous ammonium sulfate, Fe(NH4)2(SO4)2·6H2O, when dissolved in excess oxalic acid, H2C2O4 , various reacton take place are:- |
||||
2Fe 2+ |
= |
2Fe 3+ + 2e- |
(2e- "lost") |
|
H2O + HO2- + 2e- |
= |
3OH- |
(2e- "gained") |
|
H2O + HO2- + 2Fe 2+ |
= |
2Fe3+ + 3OH- |
(net reaction) |
FeC2O4 is the source of the Fe 2+ and over all reaction is
Fe(NH4)2(SO4)2· 6H2O + H2C2O4 = FeC2O4(s) + H2SO4 + (NH4)2SO4 + 6H2O(l)
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) asFe3+ + 3OH- = Fe(OH)3(s) |
|
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 |
The complexity of the series of reactions described in
equations 1 - 4 may be greatly simplified by following the
Fe2+/Fe3+ ion throughout. One discovers that
for every mole of
Fe(NH4)2(SO4)2].
6H2O used as starting material, one mole of
K3[Fe(C2O4)3].3H2O
will be obtained as the final product.
Fe(NH4)2(SO4)4. 6H2O + H2C2O4 |
= |
FeC2O4 + ......etc. |
(Fe2+ = Fe2+) |
|
FeC2O4 + K2C2O4 + H2O2 |
= |
Fe(OH)3(s) + .....etc |
(Fe2+ = Fe3+) |
|
Fe(OH)3 + H2O2 + K2C2O4 |
= |
K3[Fe(C2O4)3]·3H2O + ..etc |
(Fe3+ = Fe3+) |