Complete the conversion table below:
We will pretend that each drop of solution is 1 mL. Using your average number of drops of KMnO4 as the number of mLs (1.33 drops), use the data to calculate the molarity of the FeSO4.
___ Mn+7 + ___ Fe+2 → ___ Mn+2 ___ Fe+3
___ mLs 20 mLs
0.01 M ? M
What should the conversion table/answer look like?
Chemistry post lab question :(?
2018-05-16 8:00 am
回答 (1)
2018-05-16 8:10 am
MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe⁵⁺(aq)
Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5
Volume of MnO₄⁻ = (1.33 drops) × (1 mL/drop) = 1.33 mL
No. of milli-moles of MnO₄⁻ = (0.01 mmol/mL) × (1.33 mL) = 0.0133 mmol
No. of milli-moles of Fe²⁺ = (0.0133 mmol) × (5/1) = 0.0665 mmol
Molarity of Fe²⁺ = (0.0665 mmol) / (20 mL) = 0.00333 M
Molarity of FeSO₄ = 0.00333 M
Mole ratio MnO₄⁻ : Fe²⁺ = 1 : 5
Volume of MnO₄⁻ = (1.33 drops) × (1 mL/drop) = 1.33 mL
No. of milli-moles of MnO₄⁻ = (0.01 mmol/mL) × (1.33 mL) = 0.0133 mmol
No. of milli-moles of Fe²⁺ = (0.0133 mmol) × (5/1) = 0.0665 mmol
Molarity of Fe²⁺ = (0.0665 mmol) / (20 mL) = 0.00333 M
Molarity of FeSO₄ = 0.00333 M
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