In this experiment you will react an aqueous solution containing 3.20g of FeCl3. 6H2O with an aqueous solution containing excess K2C2O4 .H2O, assuming that all of the Fe originally in the FeCl3....

  1. In this experiment you will react an aqueous solution containing 3.20g of FeCl3. 6H2O with an aqueous solution containing excess K2C2O4 .H2O, assuming that all of the Fe originally in the FeCl3. 6H2O (moles of iron in the formula) solution ends up in the product, how many moles of product should be obtained?

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marizi's profile pic

marizi | High School Teacher | (Level 1) Associate Educator

Posted on

The two hydrates are dissolved in water to react. As a result, the water of hydration leaves. 


The reaction will be double displacement:

2FeCl3 + 3K2C2O4 ---- > Fe2(C2O4)3 + 6KCl

Conversion of g  FeCl3.6H2O to moles of FeCl3.6H2O:

3.2 g FeCl3.6H2O *    1 mol FeCl3.6 H2O

                               ------------------------------ = 0.01184 mol FeCl3.6 H2O

                                  270.32 g FeCl3.6 H2O

Conversion of mole  FeCl3.6H2O to moles of Fe2(C2O4)3:

0.01184 mol FeCl3.6 H2O *    1mol FeCl3.6                1mol Fe2(C2O4)3

                                              --------------------------- *  ----------------------- 

                                               1 mol FeCl3.6 H2O        2 mol FeCl3 

                                           =5.9 x10-3 mol Fe2(C2O4)3

d

 0.01184 mol FeCl3.6 H2O *    1mol FeCl3.6                6 mol KCl

                                              --------------------------- *  -----------------------  

                                              1 mol FeCl3.6 H2O        2 mol FeCl3 

                                        = 3.6 x10 -2 mol KCl

Products are 5.9 x10-3 mol Fe2(C2O4)3  and    3.6 x10 -2 mol KCl.

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rapunzel's profile pic

rapunzel | College Teacher | (Level 1) eNoter

Posted on

Molar mass of FeCl3.6H2O = 270.3Mass = 3.2 gMole= mass/molar mass  = 3.2/270.3 = 0.0118387 Hence mole of product formed= 0.0118387

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