Chemistry – A Mole Ratio :Reaction between iron and copper(II) sulfate

October 7, 2008 at 3:24 pm 4 comments


To observe the single replacement reaction between iron and copper(II) sulfate; calculate mole ratios; find percent yield .

Wear safety goggles and aprons
Handle chemicals (such as CuSO4) responsible and appropriately
Do not touch the hot plate with your hands

Data Table:

Data for the Reaction of Copper(II) Sulfate and Iron
Mass of empty 150-mL beaker 50.307g
Mass of 150-mL beaker + CuSO4X5H2O 63.163g
Mass of CuSO4X5H2O 12.836g
Mass of iron fillings 2.677g
Mass of 150-mL beaker and dried copper 52.760g
Mass of dried copper 2.453g
Observations Rusting, color change, copper peeling off



                 (1)159.61                                    (1) 63.55
Fe       +    CuSO4         ->      FeSO4     +      Cu
2.677g       12.836g                                         Xg
 Excess      LR

159.61         63.55    theoretical = 5.11g
12.84              X
            % Yield     =    Actual                 2.453g      X 100  =   48.00%
                                 Theoretical           5.110g 
The reason why our percent yield was so low in this experiment was because our iron sample sat at the bottom of the beaker thus not allowing the unexposed bottom half to react with the copper(II) sulfate liquid. By standing the iron up we could have easily gotten more iron to react with the copper.


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4 Comments Add your own

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  • 3. APChemStudent  |  March 15, 2018 at 10:48 pm

    this is wrong, as iron is the limiting reactant in the reaction so instead of converting copper sulfate into moles and then multiplying that by grams in one mole of copper, you should use iron instead of copper sulfate! not possible that your percent yield would be so low

    • 4. APChemStudent  |  March 15, 2018 at 10:48 pm

      sorry *the limiting reactant in the reaction is iron


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