Question

Identify the limiting reactant in the reaction of iron and chlorine to form FeCl₃, if 2.21×10¹ g of Fe and 3.59×10¹ g of Cl₂ are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.

Formula of limiting reactant =

________

Amount of excess reactant remaining =___________ g

Answer 1

           2 Fe                 +               3 Cl2                   -------------->                      2 FeCl3

2.21×10¹ g/ 55.85                   3.59×10¹ g/71

     0.3957moles                       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