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Potassium manganate(VII) concentration = 0.0020 M
Mass of two tablets = 0.980 g
Average titre = 23.3 cm3
Number of moles of manganate(VII) = 23.3/1000 × 0.0020
|
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ⇒ Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) |
MnO4- º 5Fe2+
\ Number of moles of iron(II) in 10cm3 = 5 × 23.3/1000 × 0.0020
\ Number of moles of iron(II) in 100cm3 = 10 × 5 × 23.3/1000 × 0.0020
FeSO4 = 152
\ Mass of iron(II) sulfate in 100cm3 (2 tablets) = 152 × 10 × 5 × 23.3/1000 × 0.0020
\ Mass of iron(II) sulfate in 1 tablet = ½ × 152 × 10 × 5 × 23.3/1000 × 0.0020 = 0.177 g
This is 177 mg which compares to the manufacturer's stated value of 200 mg. Have we been swindled or would the errors in the experiment tend to reduce the measured value?
Each tablet weighed 0.980/2 = 0.490 g so the percentage of iron(II) sulfate in each tablet is:
0.177/0.490 × 100 = 36.1%