Problem 80 - General Chemistry: Principles and Modern Applications

Question

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

Answer

The balanced chemical equation is: \(4FeS_{2}(s) + 15O_{2}(g) + 14H_{2}O(l) \rightarrow 4Fe^{3+}(aq) + 8SO_{4}^{2-}(aq) + 16H^{+}(aq)\). The minimum amount of \(\mathrm{CaCO}_{3}\) needed to prevent ground water contamination per kilogram of tailings S is 0.187 kg.

Step 1: 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: \(4FeS_{2}(s) + 15O_{2}(g) + 14H_{2}O(l) \rightarrow 4Fe^{3+}(aq) + 8SO_{4}^{2-}(aq) + 16H^{+}(aq)\)

Step 2: 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.

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is an essential skill in chemistry, representing the conservation of mass where the number of each type of atom on the reactants' side equals the atoms on the products' side. To balance equations, we need to accurately adjust the coefficient numbers to ensure each element complies with this law. In our exercise, we start with the unbalanced oxidation of pyrite: - Unbalanced: \[ \mathrm{FeS}_{2}(s) + \mathrm{O}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{SO}_{4}^{2-}(aq) + \mathrm{H}^{+}(aq) \] By analyzing this reaction, we break it down into half-equations. This enables us to balance the redox changes separately. The oxidation half-equation for pyrite and the reduction half for oxygen are: - Oxidation: \[ \mathrm{FeS}_{2}(s) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{SO}_{4}^{2-}(aq) \] - Reduction: \[ \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \] After meticulously balancing these half-equations to accommodate changes in electrons and combining them, the spectroscopically balanced equation becomes: - Final Balanced: \[ 4 \mathrm{FeS}_{2}(s) + 15 \mathrm{O}_{2}(g) + 14 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{Fe}^{3+}(aq) + 8 \mathrm{SO}_{4}^{2-}(aq) + 16 \mathrm{H}^{+}(aq) \] Following these steps ensures all elements are conserved, respecting the law of conservation of mass.

Environmental Chemistry

Environmental chemistry focuses on the chemical processes occurring in the environment that are influenced by human activity. A prime example from our exercise is acid mine drainage (AMD), a severe environmental issue resulting from mining activities. The tailings from iron extraction contain pyrite, which, when exposed to air and water, oxidizes and produces acidic conditions. The process:

• Pyrite (\(\mathrm{FeS}_{2}\)) oxidizes to release \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}^{+}\) ions.
• This results in the production of acidic waters that can leach into surrounding water bodies.
• The excess \(\mathrm{H}^{+}\) ions increase water acidity, harming aquatic life and affecting water quality.

To manage AMD, practices like adding limestone (\(\mathrm{CaCO}_{3}\)) are implemented to neutralize the acidity. This highlights the intersection of chemistry and environmental science by applying chemical reactions to mitigate environmental harm.

Neutralization Reactions

Neutralization reactions are chemical reactions where acids and bases react to form water and a salt, effectively reducing the acidity of the mixture. In the context of mining waste, neutralization is a critical practice to prevent environmental damage. The reaction involving limestone (\(\mathrm{CaCO}_{3}\)) is typical:- Neutralization Reaction: \[ \mathrm{CaCO}_{3}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Ca}^{2+}(aq) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g) \] This equation shows how calcium carbonate reacts with sulfuric acidity in the waste, transforming potentially harmful \(\mathrm{H}^{+}\) ions into neutral water, along with calcium ions and carbon dioxide gas. Consequently, neutralization helps to stabilize the environment in the vicinity of mining areas:

• Limestone effectively neutralizes the acidity from \(\mathrm{H}^{+}\) ions.
• The process reduces the risk of harmful acidic water affecting ecosystems.
• This approach exemplifies how chemistry can solve real-world environmental issues.

Implementing such reactions improves water safety and quality, illustrating the importance of understanding chemical principles to protect our environment from industrial impacts.