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

Select one:
a. 213
b. 252
c. 198
d. 244
Noxt page

Question

Iron tablets were analyzed using $0.05 \mathrm{M}, 18 \mathrm{~mL} \mathrm{KMnO}_{4}$. ( $\mathrm{MW}$ of iron $56 \mathrm{~g} / \mathrm{mol}$ ) Calculate the mass of iron presenting in the tables in milligrams.

Select one:
a. 213
b. 252
c. 198
d. 244
Noxt page

Accepted Answer

***C*** ∻Key Concept∻ ⚹Stoichiometry and molarity in redox reactions⚹ ∻Explanation∻ ⚹To find the mass of iron in the tablets, we need to use the stoichiometry of the reaction between $\mathrm{Fe}$ and $\mathrm{KMnO}_4$. The balanced equation for the reaction in acidic solution is: $$5 \mathrm{Fe}^{2+} + \mathrm{MnO}_4^- + 8 \mathrm{H}^+ ightarrow 5 \mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4 \mathrm{H}_2\mathrm{O}$$ This shows that 1 mole of $\mathrm{KMnO}_4$ reacts with 5 moles of $\mathrm{Fe}^{2+}$. Using the molarity and volume of $\mathrm{KMnO}_4$, we can calculate the moles of $\mathrm{Fe}^{2+}$ and then the mass of iron. Moles of $\mathrm{KMnO}_4$ used = Molarity $ imes$ Volume (in liters) = $0.05 \mathrm{M} imes 0.018 \mathrm{L} = 9 imes 10^{-4}$ moles Moles of $\mathrm{Fe}^{2+}$ = $5 imes$ Moles of $\mathrm{KMnO}_4$ = $5 imes 9 imes 10^{-4}$ moles = $4.5 imes 10^{-3}$ moles Mass of iron = Moles of $\mathrm{Fe}^{2+} imes$ Molar mass of iron = $4.5 imes 10^{-3} imes 56 \mathrm{g/mol} = 0.252 \mathrm{g}$ Since the question asks for the mass in milligrams, we convert grams to milligrams: $0.252 \mathrm{g} imes 1000 \mathrm{mg/g} = 252 \mathrm{mg}$⚹◊What is the stoichiometry of the reaction between iron and $\mathrm{KMnO}_{4}$ in this analysis?⍭ Generate me a similar question◊

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