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Posted 3 months ago
Iron tablets were analyzed using 0.05M,18 mLKMnO40.05 \mathrm{M}, 18 \mathrm{~mL} \mathrm{KMnO}_{4}. ( MW\mathrm{MW} of iron 56 g/mol56 \mathrm{~g} / \mathrm{mol} ) Calculate the mass of iron presenting in the tables in milligrams.

Select one:
a. 213
b. 252
c. 198
d. 244
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Answer from Sia
Posted 3 months ago
Key Concept
Stoichiometry and molarity in redox reactions
To find the mass of iron in the tablets, we need to use the stoichiometry of the reaction between Fe\mathrm{Fe} and KMnO4\mathrm{KMnO}_4. The balanced equation for the reaction in acidic solution is: 5Fe2++MnO4+8H+5Fe3++Mn2++4H2O5 \mathrm{Fe}^{2+} + \mathrm{MnO}_4^- + 8 \mathrm{H}^+ \rightarrow 5 \mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4 \mathrm{H}_2\mathrm{O} This shows that 1 mole of KMnO4\mathrm{KMnO}_4 reacts with 5 moles of Fe2+\mathrm{Fe}^{2+}. Using the molarity and volume of KMnO4\mathrm{KMnO}_4, we can calculate the moles of Fe2+\mathrm{Fe}^{2+} and then the mass of iron. Moles of KMnO4\mathrm{KMnO}_4 used = Molarity ×\times Volume (in liters) = 0.05M×0.018L=9×1040.05 \mathrm{M} \times 0.018 \mathrm{L} = 9 \times 10^{-4} moles Moles of Fe2+\mathrm{Fe}^{2+} = 5×5 \times Moles of KMnO4\mathrm{KMnO}_4 = 5×9×1045 \times 9 \times 10^{-4} moles = 4.5×1034.5 \times 10^{-3} moles Mass of iron = Moles of Fe2+×\mathrm{Fe}^{2+} \times Molar mass of iron = 4.5×103×56g/mol=0.252g4.5 \times 10^{-3} \times 56 \mathrm{g/mol} = 0.252 \mathrm{g} Since the question asks for the mass in milligrams, we convert grams to milligrams: 0.252g×1000mg/g=252mg0.252 \mathrm{g} \times 1000 \mathrm{mg/g} = 252 \mathrm{mg}

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