Question

From Part A: Mass of KyFe(C204)y · 2H20 prepared : 4.450 g Mass of FeCl3 : 1.60 g From Part B: % Potassium in compound : 10.70 % % Iron (from ion exchange & titration vs. NaOH) : 12.40 % From Part C: % Oxlate : 49.14 %

Answer 1

As complex formed between iron and oxalate ion having following formula in which iron present in +3 oxidation state

Q1. Answer

K3​​​​​3[Fe(C​​​​​​2​​C4)3]. 3 H​​2O

Here K=3=x

C2O4= 3 =y

H2O=3=z

Q.2 According to stoichiometry only 1 mole i.e. 162.2 gm/mol of FeCl3 is required

Therefore moles of FeCl3 = 1.6 X 1/162.2

=0.009864 molesF

Q.3. FeCl3+ 3 K2C2O4 → K3[Fe(C2O4)3] +3KCl

According to equation 1 mole complex formed i.e. 491.25 gm / mole

Hence theroetically 1 mole complex is formed

Q.4

Actual moles of complex formed

=Given mass of complex/molar mass of complex

=4.450/491.25

=0.00905 moles

Q.5

% yeild of complex= actual yeild in grams/theroetical yeild in gm X 100

=4.450/4.9125 X 100

=90.584%