Question

A student performed the experiment by using 2.6415 g of FeCl3•6H2O and 7.6211 g K2C2O4•H2O. Determine the limiting reactant when the K3Fe(C2O4)3•3H2O(s) product is formed as per the reaction given in the lab instructions.

Answer 1

first write the balanced equation

FeCl3 ·6H2O + 3 K2C2O4 ·H2O = K3[Fe(C2O4)3]·3H2O + 3 KCl + 6 H2O

according to reaction to get one mole of K3[Fe(C2O4)3]·3H2O

we need one mole of FeCl3 ·6H2O and 3 moles of K2C2O4 ·H2O

so no moles of FeCl3•6H2O = weight /molar mass

= 2.6145 / 270.296

= 0.009672 moles

no of moles of K2C2O4•H2O

= 7.6211 / 184.231

= 0.04136 moles

for 0.009672 moles of FeCl3•6H2O required 0.029 moles of K2C2O4•H2O

but we have 0.04136 moles that means

FeCl3•6H2O is limiting agent.