The mass of FeCl3•6H2O is 2.6033 and the mass of K2C2O4•H2O is 6.256. Which of these is the limiting reactant ?​?

Molar mass of FeCl₃•6H₂O = (55.8 + 35.5×3 + 1.0×12 + 16.0×6) g/mol = 270.3 g/mol
Molar mass of K₂C₂O₄•H₂O = (39.0×2 + 12.0×2 + 16.0×5 + 1.0×2) g/mol = 184.0 g/mol

This is a precipitation reaction, and the equation is :
2 FeCl₃(aq) + 3 K₂C₂O₄(aq) → Fe₂(C₂O₄)₃(s) + 6 KCl(aq)
OR: Mole ratio FeCl₃ : K₂C₂O₄ = 2 : 3

Initial no. of moles of FeCl₃ = Initial no. of moles of FeCl₃•6H₂O = (2.6033 g) / (270.3 g/mol) = 0.009631 mol
Initial no. of moles of K₂C₂O₄•H₂O = Initial no. of moles of K₂C₂O₄•H₂O = (6.256 g) / (184.0 g/mol) = 0.03400 mol

When 0.009631 mol of FeCl₃•6H₂O completely reacts :
No. of moles of K₂C₂O₄•H₂O needed = (0.009631 mol) × (3/2) = 0.01445 mol < 0.03400 mol
Hence, K₂C₂O₄•H₂O is in excess, and FeCl₃•6H₂O is the limiting reactant.

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