A reaction of 7.20 mL of iron?
A reaction of 7.20 mL of Iron(III) chloride solution (0.40 g/mL) and 8.95 g of K2C2O4.H2O (Potassium oxalate monohydrate) produced 6.22 g of K3Fe(C2O4)3.3H2O:
Calculate number moles of Iron(III) chloride?Calculate number moles K2C2O4.H2O (Potassium oxalate monohydrate). (Answer round off to 4 decimal places, and the unit is molesCalculate the percent yield of the synthesis?
- Roger the MoleLv 74 weeks ago
(0.00720 L FeCl3) x (0.40 g/mL) / (162.204 g FeCl3/mol) = 0.017755 mol =
0.0178 mol FeCl3
(8.95 g K2C2O4·H2O) / (184.2309 g K2C2O4·H2O/mol) = 0.048580 mol =
0.0486 mol K2C2O4·H2O
FeCl3 + 3 K2C2O4 → K3Fe(C2O4)3 + 3 KCl
0.0486 mole of K2C2O4 would react completely with 0.0486 x (1/3) = 0.0162 mole of FeCl3, but there is more FeCl3 present than that, so FeCl3 is in excess and K2C2O4 is the limiting reactant.
(0.048580 mol K2C2O4) x (1 mol K3Fe(C2O4)3 / 3 mol K2C2O4) x
(491.2427 g K3Fe(C2O4)3·3H2O/mol) = 7.95486 g K3Fe(C2O4)3·3H2O in theory
(6.22 g actual) / (7.95486 g theoretical) = 0.7819 = 78.2% yield K3Fe(C2O4)3·3H2O