Partner Data Sheet Mass of FeClh.6 H20 (8) Mass of CHsCo2Na. 3 H20 (8) Volume of
ID: 1026308 • Letter: P
Question
Partner Data Sheet Mass of FeClh.6 H20 (8) Mass of CHsCo2Na. 3 H20 (8) Volume of 2,4-pentanedione (mL) Mols of FeCls 6 H0 580 0.00 S62szmois 0.60 50 2 52 mois 270.2157 g Feci3.lozo Feclz 6Hzo Mols of CH CO2Na.3 H20 1120 Mols of 2,4-pentanedione (look up density) +120 100.139 Theoretical yield of Fe (C11:0(mol) Theoretical yield of Fe (CsH70)18) Mass of beaker (g) Mass of beaker plus Fe(CsH 02) (g) Actual Mass of Fe(CsH,02)s (8) Percent yield of Fe Fe(CsH,0)3 (mol) 177 C star 152 end compiete Melting point (oc) weite Exp 4Explanation / Answer
Ans. Balanced reaction:
FeCl3.6H2O (aq) + 3 C5H8O2 + 3 CH3COONa.3H2O (aq) ------>
Fe(C5H7O2)3(s) + 3 NaCl (aq)+ 3 CH3COOH + 9 H2O (l)
# Following stoichiometry, the theoretical molar ratio of reactants-
FeCl3.6H2O : 3 C5H8O2 : 3 CH3COONa.3H2O = 1 : 3 : 3
# Now, experimental molar ratio of reactants-
FeCl3.6H2O : 3 C5H8O2 : 3 CH3COONa.3H2O
= 0.0050252 mol : 0.01865 mol : 0.0195 mol
= 1 : 3.7 : 3.8
# Comparing the theoretical and experimental molar rations of reactants, the experimental moles of C5H8O2 and CH3COONa.3H2O are greater than their theoretical value of 3 moles while that of FeCl3.6H2O is kept constant at 1 mol. Therefore, C5H8O2 and CH3COONa.3H2O are reagents in excess, and FeCl3.6H2O is the limiting reactant.
# Yield: The formation of product follows the stoichiometry of limiting reactant.
Following stoichiometry, 1 mol FeCl3.6H2O produces 1 mol Fe(C5H7O2)3.
So,
Moles of Fe(C5H7O2)3 produced = 0.0050252 mol = moles of limiting reactant
And,
Theoretical mass of Fe(C5H7O2)3 produced = theoretical moles x MW
= 0.0050252 mol x (353.17514 g/ mol)
= 1.7748 g
Therefore, theoretical yield of Fe(C5H7O2)3 = 1.7748 g
# Now, % yield = (Actual yield / Theoretical yield) x 100
= (.1.0204 g / 1.7748 g) x 100
= 57.494 %