Sunday, July 10, 2005
Synthesis of bimetallic nanoparticles of Co & Fe particles
Loading = 10% cobalt (II) & 10% iron (III)
Ratio between Cobalt and Iron = 1: 1
Base= 1g of silica
Reducing agent = Sodium borohydride
80% - 1g
1% - (1/80)g
20% - 0.250g of catalyst
let x be the mols of Iron(III)
0.250g = (55.85x) + (58.93x)
0.250 = 114.78x
x = 2.1780798 X 10-3 mols of FE
since ratio is 1: 1 ,
therefore the mols of Co = (8.854886126 X 10-4 )X 4
= 2.1780798 X 10-3 mols
Total mols of catalyst = mols of Co + mols of Fe
= 4.35615961 X 10-3 mols
Amt of Fe(NO3)3 = mols of Fe X Mr of iron(III) nitrate
=mols of Fe X 404
=0.879944241g
Amt of Co(NO3)2 = mols of Co X Mr of Cobalt(II) nitrate
= mols of Co X 291.03
= 0.633886565g
Since total amt of catalyst to be reduced = 4.287135165 X 10-3 mols
Thus At least. = 4.35615961X 10-3 mols of reducing agent is needed.
Since we are using 0.1M of reducing agent
Amt of h20 needed to prepare reagent = total mols of catalyst / 0.1M X 1000 X 10
= (4.35615961X 10-3 mols)/0.1M X 1000 X10
= 435.6159961ml of water is needed
BOYAKASHA!!!!! haha!! Eat your heart out!!
