********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.

Explanation / Answer

2 FeC2O4*2H2O(s) + 3 K2C2O4*H2O(aq) + H2O2(aq) + H2C2O4(aq) -->2 K3Fe(C2O4)3*3H2O(s) + 3 H2O(l)

Fe is going from a +2 oxidation state to a +3 oxidation state which means it is oxidized.
Fe+2 --> Fe+3 + 1 e-
At the same time, the Oxygen in hydrogen peroxide is being reduced as its oxidation state goes from -1 to -2.
H2O2 + 2 e- + 2 H+ --> 2 H2O
Since we need to have two electrons lost to provide the 2 electrons gained by the Oxygen we multiply the oxidation 1/2 reaction by 2.
2 Fe+2 --> 2 Fe+3 + 2 e-

The two 1/2 reactions tell us that the Fe+2 and H2O2 MUST be in a 2:1 ratio. Adding the 2 1/2 reactions gives us:
2 Fe+2 + H2O2 + 2 H+ --> 2 Fe+3 + 2 H2O
Add 2 OH- to both sides to create water from the 2H+ ions. Then cancel the 2 water molecules which appear both as reactants and products.
2 Fe+2 + H2O2 --> 2 Fe+3 + 2OH-
Add the oxalate, water and K+ ions to both sides to make the hydrate on the product side:
6K+ + 6 C2O4-2 + 2 Fe+2 + H2O2 + 3 H2O --> 2 K3Fe(C2O4)3*3 H2O + 2OH-

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