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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Sektor404 Sektor404
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9 years ago
Fe2(SO4)3 + 3 BaC2O4 + 3 H2O +  3 K2C2O4*H2O → 2 K3[Fe(C2H4)3]*3H2O + 3 BaSO4

Ferric Sulfate = 2.50 g/( 399.88 g/mol) = 0.00625 mol
Barium Oxalate = 4.98 g/(225.346 g/mol) = 0.0221 mol
Potassium Oxalate = 2.70 g/(184.23 g/mol) = 0.0147 mol

Reactant ratio = 1:3:3, therefore potassium oxalate is the limiting reagent
The ratio of the K3[Fe(C2H4)3]*3H2O product to the K2C2O4*H2O reactant is 2:3. Thus the theoretical yield is:
2/3 x 0.0147 mol x 491.25 g/mol = 4.81 g


Does this seem correct or have I done something wrong???  Confounded Face
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