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Chapter 21: Problem 37

37\. Iodine can be prepared by allowing an aqueous solution of hydrogen iodide to react with manganese dioxide, \(\mathrm{MnO}_{2}\). The reaction is $$2 \mathrm{I}^{-}(a q)+4 \mathrm{H}^{+}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}(s)$$ If an excess of hydrogen iodide is added to \(0.200 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\), how many grams of iodine are obtained, assuming \(100 \%\) yield?

Short Answer

Expert verified
Answer: The mass of iodine obtained is 0.583 g.

Step by step solution

01

Convert the mass of MnO2 to moles

To perform stoichiometry calculations, we need to convert the mass of the reactant, manganese dioxide (MnO2), into moles. We do this by dividing the mass by the molar mass of MnO2: Molar mass of MnO2 = 54.94 (Mn) + 2 * 16.00 (O) = 86.94 g/mol Moles of MnO2 = (mass of MnO2) / (molar mass of MnO2) Moles of MnO2 = 0.200 g / 86.94 g/mol = 0.00230 mol
02

Use the stoichiometry of the reaction to determine moles of I2

From the balanced chemical equation, we can see that one mole of MnO2 reacts to produce one mole of I2. Therefore, if we have 0.00230 moles of MnO2, we will also have 0.00230 moles of I2 produced.
03

Convert moles of I2 to mass

Now we need to convert moles of iodine (I2) back into mass. We do this by multiplying the moles by the molar mass of I2: Molar mass of I2 = 2 * 126.90 g/mol = 253.80 g/mol Mass of I2 = (moles of I2) * (molar mass of I2) Mass of I2 = 0.00230 mol * 253.80 g/mol = 0.583 g
04

Final answer

Assuming a 100% yield, the mass of iodine obtained from a 0.200 g sample of manganese dioxide (MnO2) would be 0.583 g of I2.

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Most popular questions from this chapter

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