chem lab question

[匿名]

2011-01-16 9:38 am

A 25.00 cm^3 of an unknow iron(II) sulphate solution, was made acidic with dilute suphuric acid and titrated with standard 0.03247 M potassium peranganate solution. The titration required 40.38cm^3 of potassium permanganate solution. Calculate the concentratin of the unknown iron (II) sulphate soultion in g/L.

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不用客氣

2011-01-16 6:41 pm

✔ 最佳答案

A 25.00 cm³ of an unknown iron(II) sulphate solution, was made acidic with dilute suphuric acid and titrated with standard 0.03247 M potassium peranganate solution. The titration required 40.38 cm³ of potassium permanganate solution. Calculate the concentration of the unknown iron(II) sulphate solution in g/L.

5Fe²⁺(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H2O(l)
Mole ratio Fe²⁺ : MnO4⁻ = 5 : 1

No. of moles of KMnO4 = 0.03247 x (40.38/1000) = 0.001311 mol
No. of moles of MnO4⁻ = 0.001311 mol
No. of moles of Fe²⁺ = 0.001311 x 5 = 0.006555 mol
No. of moles of FeSO4 = 0.006555 mol

Molar mass of FeSO4 = 55.85 + 32.06 + 16x4 = 151.91 g/mol
Mass of FeSO4 = 0.006555 x 151.91 = 0.9958 g
Concentration of FeSO4 = 0.9958 / (25/1000) = 39.83 g/L

參考: micatkie

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