1. The chemical equation for the synthesis, balanced only for the reactants, is

Question

1.  The chemical equation for the synthesis, balanced only for the reactants,  is as follows.  Supply the coefficients for KCl and H2O required to balance  the equation completely.

 FeCl3·6H2O(aq)  +  3K2C2O4·H2O(aq)  

 ®  K3Fe(C2O4)3·3H2O(s)   + ____KCl(aq)   + ____H2O(l)

 Note that the reactants are used as their hydrates.  FeCl3·6H2O means  that one mol of the compound corresponds to one mol of FeCl3 and six  mol of water.  K2C2O4·H2O means that three mol of the compound  corresponds to three mol K2C2O4 and three mol of water.

2. Calculate the amount (mols) corresponding to 5.00 g of FeCl3·6H2O  (270.3 g/mol) and to 11.00 g of K2C2O4·H2O (184.2 g/mol).   Note that the  molar masses necessarily include the water of hydration.
 

3.  On the basis of the chemical amounts calculated in no. 2 and the balanced  chemical equation, determine the limiting reactant showing your  calculations and telling why you chose that reactant.

4.  Calculate the theoretical yield of the coordination compound (molar mass  = 491.3 g/mol).

The product, like the reactants, is a hydrate.  So the molar mass given is for the hydrate.

5. Calculate the percent yield based on an (assumed) actual yield of 7.52 g.

Explanation / Answer

1) FeCl3 . 6H2O + 3K2C2O4. H2O -----------------> K3[Fe(C2O4)3].3H2O +3KCl + 6H2O
2) mol of  FeCl3 . 6H2O = 5 gm / 270.3 g/mol = 0.01849 mol
   mol of K2C2O4·H2O = 11 gm / 184.2 g/mol = 0.05971 mol
3)Determining Limiting reactant
0.01849 mol FeCl3.6H2O * ( 1 mol rxn/1 mol FeCl3.6H2O) = 0.01849 mol FeCl3.6H2O
0.05971 mol K2C2O4·H2O *(1 mol rxn/3 mol K2C2O4·H2O) = 0.019903 mol K2C2O4·H2O
FeCl3.6H2O is the limiting reactant ( less number)
4) 0.01849 mol FeCl3.6H2O * 491.3 g/mol = 9.084137 gms
Theoretical yield = 9.084137 gms
5) Percent yield = (actual yield / theoretical yield)*100
   % yield = (7.52 gm / 9.084137 gm)*100 = 82.78 %

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