Name___________________________

Lab Section________

Chem 161U: Prelab Exercise #1 - Due at the beginning of lecture - June 24, 1996

No Late Homeworks Accepted

Step 1: Fe(NH4)2(SO4)2.6H2O + H2C2O4 --> FeC2O4.2H2O + Spectator ions

Step 2: 2FeC2O4.2H2O + 3K2C2O4 + H2O2 + H2C2O4 --> 2K3Fe(C2O4)3.3H2O

Fe(NH4)2(SO4)2.6H2O = 392.2 g .mol-1
H2C2O4 = 90.0 g. mol-1
FeC2O4.2H2O = 179.9 g.mole-1
K2C2O4 = 162.2 g.mol-1
H2O2 = 34.0 g.mol-1
K3Fe(C2O4)3.3H2O = 491.3 g.mol-1

1. a) Why do we need an oxidizing agent in our synthesis of Fe(C2O4)33- ? (1 pt.)

b) Name the oxidizing agent and the species that is oxidized. (2 pts)

oxidizing agent _________, species oxidized __________

2. a) Assuming ferrous ammonium sulfate is the limiting reagent, calculate the theoretical yield if you start with 3.0 g of ferrous ammonium sulfate hexahydrate. (2 pts)

b) Calculate the percentage yield if 2.3 g of K3Fe(C2O4)3.3H2O is obtained from 3.0 g of Fe(NH4)2(SO4)2.6H2O (3 pts)

c) Beginning with the two half reactions:

2FeC2O4.2H2O + 3K2C2O4 + H2C2O4 + H2O --> ___K3Fe(C2O4)3.3H2O + ___e- + ___H+


H2O2 + 2H+ +2e- ----> ___H2O

Balance the above redox half reactions and then generate the balanced reaction in step 2. (2 pts)