2
Fe               Â
+Â Â Â Â Â Â Â Â Â Â Â Â Â Â
3
Cl2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
-------------->Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
2 FeCl3
2.21×101 g/
55.85Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
3.59×101 g/71
   Â
0.3957moles                     Â
0.5056 moles
one mole Fe needs 1.5mole Cl2 for complete reaction
0.3957mole Fe needs 0.3957*1.5 = 0.59355moles of Cl2
So here the limiting reagent is Cl2 - chlorine
1.5 mole Cl2 needs one mole of Fe
0.5056moles of Cl2 needs 0.5056/1.5= 0.337 mole of Fe
So excess reagent = 0.3957-0.3370 = 0.0587moles of iron = 3.2784
grams
Weight = moles * molecular weight = 0.0587*55.85 =
3.2784grams
