If 24.38ml of KMnO4 solution are required to oxidize 25ml of 0.03156M of Na2C2O4 solution, what is the concentration of KMnO4?
A certain brand of iron supplements contains FeSO4(7h2o) with miscellaneous binders and fillers. suppose 21.32ml of KMnO4 solution (from question above) are needed to oxidize Fe2+ to Fe3+ in a O.5580 g pill. What is the mass percent of FeSO4(7H2O) in the pill? ( molar mass 278.03g/mol)
2MnO4- + 5C2O4^2- + 16H+ --> 2Mn^2+ + 10CO2 + 8H2O
How to calculate concentration and mass percent?
2017-05-25 11:41 pm
回答 (1)
2017-05-26 5:20 am
2 KMnO4 + 5 Na2C2O4 + 16 H{+} → 2 Mn{2+} + 8 H2O + 10 CO2 + 2 K{+} + 10 Na{+}
(25 mL) x (0.03156 M Na2C2O4) x (2 mol KMnO4 / 5 mol Na2C2O4) / (24.38 mL KMnO4) = 0.012945 = 0.013 M KMnO4
KMnO4 + 5 Fe{2+} + 8 H{+} → 5 Fe{3+} + Mn{2+} + 4 H2O + K{+}
(0.012945 mol/L KMnO4) x (0.02132 L KMnO4) x (5 mol Fe / 1 mol KMnO4) x (1 mol FeSO4·7H2O / 1 mol Fe) x
(278.03 g/mol) / (0.5580 g) = 0.687569 = 68.76% FeSO4·7H2O by mass
(25 mL) x (0.03156 M Na2C2O4) x (2 mol KMnO4 / 5 mol Na2C2O4) / (24.38 mL KMnO4) = 0.012945 = 0.013 M KMnO4
KMnO4 + 5 Fe{2+} + 8 H{+} → 5 Fe{3+} + Mn{2+} + 4 H2O + K{+}
(0.012945 mol/L KMnO4) x (0.02132 L KMnO4) x (5 mol Fe / 1 mol KMnO4) x (1 mol FeSO4·7H2O / 1 mol Fe) x
(278.03 g/mol) / (0.5580 g) = 0.687569 = 68.76% FeSO4·7H2O by mass
收錄日期: 2021-04-24 00:26:19
原文連結 [永久失效]:
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