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Chapter 5: Problem 65

An iron ore sample weighing \(0.9132 \mathrm{g}\) is dissolved in \(\mathrm{HCl}(\mathrm{aq}),\) and the iron is obtained as \(\mathrm{Fe}^{2+}(\mathrm{aq}) .\) This solution is then titrated with \(28.72 \mathrm{mL}\) of \(0.05051 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) What is the mass percent Fe in the ore sample? \(6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow_{6 \mathrm{Fe}^{3+}}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
The mass percent of Fe in the ore sample is 53.2%.

Step by step solution

01

Calculate the number of moles of Cr2O72- used in the titration

This can be obtained from the molarity and volume of the K2Cr2O7 solution. The molarity is 0.05051 M, and the volume is 28.72 mL or 0.02872 L. Therefore, the number of moles of Cr2O72- is \(0.05051 M \times 0.02872 L = 0.00145 moles\).
02

Calculate the number of moles of Fe2+

From the stoichiometry of the reaction, 6 moles of Fe2+ react with 1 mole of Cr2O72-. Therefore, the number of moles of Fe2+ is \(0.00145 moles \times 6 = 0.0087 moles\).
03

Calculate the mass of iron

This can be obtained by multiplying the number of moles of Fe2+ (found in step 2) by the molar mass of iron. The molar mass of iron is 55.845 g/mol. Therefore, the mass of iron is \(0.0087 moles \times 55.845 g/mol = 0.486 g\).
04

Calculate the mass percent of iron

This can be obtained by dividing the mass of iron (found in step 3) by the mass of the sample and multiplying by 100. The mass of the sample is 0.9132 g. Therefore, the mass percent of Fe is \((0.486 g / 0.9132 g) \times 100 = 53.2% \).

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Most popular questions from this chapter

How many milliliters of 2.155 M KOH are required to titrate \(25.00 \mathrm{mL}\) of \(0.3057 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) (prop-ionic acid)?

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?

How many milliliters of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) are required to titrate \(50.00 \mathrm{mL}\) of \(0.0526 \mathrm{M} \mathrm{HNO}_{3} ?\)

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

When treated with dilute \(\mathrm{HCl}(\mathrm{aq}),\) the solid that reacts to produce a gas is (a) \(\mathrm{BaSO}_{3} ;\) (b) \(\mathrm{ZnO};\) (c) \(\mathrm{NaBr} ;\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\).
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