Iron (Fe) is obtained from rock that is extracted from open pit mines and then crushed. The process used to obtain the pure metal from the crushed rock produces solid waste, called tailings, which are stored in disposal areas near the mines. The tailings pose a serious environmental risk because they contain sulfides, such as pyrite ( \(\mathrm{FeS}_{2}\) ), which oxidize in air to produce metal ions and \(\mathrm{H}^{+}\) ions that can enter into surface water or ground water. The oxidation of \(\mathrm{FeS}_{2}\) to \(\mathrm{Fe}^{3+}\) is described by the unbalanced chemical equation below. \(\mathrm{FeS}_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\quad \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \quad(\text { not balanced })\) Thus, the oxidation of pyrite produces \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}^{+}\) ions that can leach into surface or ground water. The leaching of \(\mathrm{H}^{+}\) ions causes the water to become very acidic. To prevent acidification of nearby ground or surface water, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is added to the tailings to neutralize the \(\mathrm{H}^{+}\) ions: \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \underset{\mathrm{Ca}^{2+}}{\longrightarrow}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Balance the equation above for the reaction of \(\mathrm{FeS}_{2}\) and \(\mathrm{O}_{2}\). [ Hint: Start with the half-equations \(\mathrm{FeS}_{2}(\mathrm{s}) \rightarrow\) \(\left.\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \text { and } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) .\right]\) (b) What is the minimum amount of \(\mathrm{CaCO}_{3}(\mathrm{s})\) required, per kilogram of tailings, to prevent contamination if the tailings contain \(3 \%\) S by mass? Assume that all the sulfur in the tailings is in the form \(\mathrm{FeS}_{2}\).

Step by step solution

01

Balance the Chemical Equation

First, we start with the half-reactions given: \(\mathrm{FeS}_{2}(s) \rightarrow \left.\mathrm{Fe}^{3+}(aq)+ \mathrm{SO}_{4}^{2-}(aq)\) and \(\mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2}O(l)\). These half-reactions need to be carefully balanced. The resulting balanced chemical equation becomes: \(4FeS_{2}(s) + 15O_{2}(g) + 14H_{2}O(l) \rightarrow 4Fe^{3+}(aq) + 8SO_{4}^{2-}(aq) + 16H^{+}(aq)\)

02

Determine the Amount of \(\mathrm{CaCO}_{3}\) Required

Now, we move on to solve part (b) of the exercise. Based on the balanced equation, each \(\mathrm{FeS}_{2}\) molecule produces 4 \(\mathrm{H}^{+}\) ions. Hence, 1 mole of \(\mathrm{FeS}_{2}\) would require 4 moles of \(\mathrm{CaCO}_{3}\) for complete neutralization. To determine the amount in kilograms, we will first calculate the number of moles of \(\mathrm{FeS}_{2}\) in 1 kg of tailings. Given that the tailings contain 3% S by mass, and all the sulfur is in the form of \(\mathrm{FeS}_{2}\), this equals to 0.03 kg of S, which equals to 0.03 kg / 32.065 g/mol (molar mass of sulfur) = 0.935 moles of S.Since 1 mole of \(\mathrm{FeS}_{2}\) contains 2 moles of S, there would be 0.935/2 = 0.468 moles of \(\mathrm{FeS}_{2}\) in 1 kg of tailings. Hence, the amount of \(\mathrm{CaCO}_{3}\) needed would be 4*0.468 moles = 1.87 moles. Converting this to mass, using the molar mass of \(\mathrm{CaCO}_{3}\) (100.09 g/mol), the result is 1.87 moles * 100.09 g/mol / 1000 g/kg = 0.187 kg of \(\mathrm{CaCO}_{3}\) per kg of tailings.

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