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Chapter 4: Problem 107

Copper exists in the form of \(\mathrm{CuFeS}_{2}\) in copper ore. Copper is isolated in a two-step process. First, CuFeS \(_{2}\) is heated with \(\mathrm{SiO}_{2}\) in the presence of oxygen to form copper(I) sulfide, \(\mathrm{CuS:} 2 \mathrm{CuFeS}_{2}+2 \mathrm{SiO}_{2}(s)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{~S}(s)+\) \(2 \mathrm{FeSiO}_{3}(s)+3 \mathrm{SO}_{2}(g) . \mathrm{Cu}_{2} \mathrm{~S}\) is then heated with oxygen to form copper and \(\mathrm{SO}_{2}(g)\). (a) Write the balanced chemical equation for the second reaction. (b) Which atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced? (c) How many grams of copper would be isolated from \(85.36 \mathrm{~g}\) of \(\mathrm{CuFeS}_{2}\) in copper ore?

Short Answer

Expert verified
The balanced chemical equation for the second reaction is: 2 Cu鈧係(s) + 3 O鈧�(g) 鈫� 4 Cu(s) + 2 SO鈧�(g). In this process, Cu is being reduced, while Fe and S are being oxidized. From 85.36 g of CuFeS鈧� in copper ore, 59.13 g of copper can be isolated.

Step by step solution

01

Write the balanced chemical equation for the second reaction

We are given that Cu鈧係 is heated with oxygen to form copper and SO鈧�. To write the balanced chemical equation, we need to balance all atoms on both sides of the equation. Since there are 2 copper atoms and 1 sulfur atom on the left-hand side, we require 2 copper atoms and 1 sulfur atom on the right-hand side as well: 2 Cu鈧係(s) + 3 O鈧�(g) 鈫� 4 Cu(s) + 2 SO鈧�(g)
02

Identify the atoms being oxidized and reduced

In the first reaction, CuFeS鈧� is heated with SiO鈧� in the presence of oxygen to form Cu鈧係, FeSiO鈧�, and SO鈧�. The species being oxidized and reduced can be identified by analyzing the change in their oxidation states: - Cu: goes from +1 in CuFeS鈧� to 0 in Cu (reduced) - Fe: goes from +2 in CuFeS鈧� to +3 in FeSiO鈧� (oxidized) - S: goes from -2 in CuFeS鈧� to +4 in SO鈧� (oxidized) Thus, Cu is being reduced, while Fe and S are being oxidized.
03

Calculate the amount of copper isolated

We are given that 85.36 g of CuFeS鈧� is used in copper ore. To determine the amount of copper isolated, we need to convert the mass of CuFeS鈧� to mole and then use stoichiometry to convert moles of CuFeS鈧� to moles of copper. Finally, we will convert moles of copper back to mass (grams). 1. Calculate the molar mass of CuFeS鈧�: M(CuFeS鈧�) = Atomic Mass of Cu + Atomic Mass of Fe + 2 * Atomic Mass of S M(CuFeS鈧�) = 63.55 g/mol + 55.85 g/mol + 2 * 32.07 g/mol = 183.54 g/mol 2. Convert mass of CuFeS鈧� to moles: moles of CuFeS鈧� = (mass of CuFeS鈧�) / (molar mass of CuFeS鈧�) moles of CuFeS鈧� = 85.36 g / 183.54 g/mol = 0.4649 mol 3. Convert moles of CuFeS鈧� to moles of Cu using stoichiometry (from the balanced chemical equation in Step 1): moles of Cu = 2 * moles of CuFeS鈧� (since coefficent of Cu鈧係 is 2 and coefficent of Cu is 4 in the balanced chemical equation) moles of Cu = 2 * 0.4649 mol = 0.9298 mol 4. Convert moles of Cu back to mass in grams: mass of Cu = moles of Cu * Atomic Mass of Cu mass of Cu = 0.9298 mol * 63.55 g/mol = 59.13 g Hence, 59.13 g of copper would be isolated from 85.36 g of CuFeS鈧� in copper ore.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
In the realm of chemistry, a chemical reaction is a process where substances, called reactants, transform into different substances known as products. In the context of copper extraction, chemical reactions are pivotal. The initial chemical equations given in the problem demonstrate the transformation of copper in its ore form \(\text{CuFeS}_{2}\) to eventually refine pure copper metal. Here, different reactants like \(\text{SiO}_{2}\) and \(\text{O}_{2}\) partake in this conversion.
  • First Reaction: Involves heating \(\text{CuFeS}_{2}\) with \(\text{SiO}_{2}\) and \(\text{O}_{2}\) to form \(\text{Cu}_{2}\text{S}\).
  • Second Reaction: \(\text{Cu}_{2}\text{S}\) is heated with \(\text{O}_{2}\) resulting in copper and \(\text{SO}_{2}\).
To balance these reactions, ensure each type of atom appears equally on both sides of the reaction arrow. This conservation of mass is crucial for the proper understanding of the reaction dynamics. Visualizing these changes, a balanced equation ensures that chemical identity and reaction stoichiometry are accurately represented.
Stoichiometry
Stoichiometry involves the calculations of reactants and products in chemical reactions. It is a powerful tool to predict yields and guide experimentations.
In the context of copper extraction:
  • The initial ore \(\text{CuFeS}_{2}\) yields copper through a series of reactions where stoichiometric ratios determine how much copper can be extracted from a given ore mass.
  • Stoichiometry allows us to convert grams of \(\text{CuFeS}_{2}\) to moles using its molar mass and then relate these moles to moles of copper produced using the balanced equation.
For instance, if 85.36 g of \(\text{CuFeS}_{2}\) is available, stoichiometry helps us calculate how many grams of copper can be isolated. This involves converting between mass and moles, using the equation coefficients to relate reactants and products.
Oxidation and Reduction
Oxidation and reduction (redox) reactions are key aspects of many chemical processes, including copper extraction.
  • Oxidation involves the loss of electrons, while reduction involves the gain of electrons.
  • In the given context, copper \(\text{(Cu)}\) is reduced from +1 in \(\text{CuFeS}_{2}\) to 0 in elemental copper, indicating it gains electrons.
  • Conversely, iron \(\text{(Fe)}\) is oxidized from +2 in \(\text{CuFeS}_{2}\) to +3 in \(\text{FeSiO}_{3}\).
  • Sulfur \(\text{(S)}\) also undergoes oxidation from -2 in \(\text{CuFeS}_{2}\) to +4 in \(\text{SO}_{2}\).
It's important to grasp how oxidation states change during the reaction. This helps identify the substance that acts as a reducing agent (provides electrons) and the oxidizing agent (accepts electrons), which is essential for understanding the chemical mechanism behind copper extraction.
Molar Mass Calculation
Calculating molar mass is vital for converting between mass and moles, a foundational step in stoichiometric calculations.
  • The molar mass of a compound is the sum of the atomic masses of its constituent elements in appropriate proportions.
  • For \(\text{CuFeS}_{2}\):
    • Copper (Cu): 63.55 g/mol
    • Iron (Fe): 55.85 g/mol
    • Sulfur (S): 32.07 g/mol (two sulfur atoms are present, so \(2 \times 32.07\) g/mol)
  • Total molar mass: \(63.55 + 55.85 + 2 \times 32.07 = 183.54 \text{ g/mol}\)
Using the molar mass allows direct conversion of grams to moles, setting the stage for further stoichiometric analysis. This precise conversion ensures accuracy when predicting the amount of product formed in a reaction鈥攍ike determining the mass of copper extracted from an ore sample.

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Most popular questions from this chapter

Bronze is a solid solution of \(\mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Sn}(s) ;\) solutions of metals like this that are solids are called alloys. There is a range of compositions over which the solution is considered a bronze. Bronzes are stronger and harder than either copper or tin alone. (a) A \(100.0-\mathrm{g}\) sample of a certain bronze is \(90.0 \%\) copper by mass and \(10.0 \%\) tin. Which metal can be called the solvent, and which the solute? (b) Based on part (a), calculate the concentration of the solute metal in the alloy in units of molarity, assuming a density of \(7.9 \mathrm{~g} / \mathrm{cm}^{3} .\) (c) Suggest a reaction that you could do to remove all the tin from this bronze to leave a pure copper sample. Justify your reasoning.

You know that an unlabeled bottle contains an aqueous solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2},\) or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} . \mathrm{A}\) friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with NaCl solutions. According to your friend's logic, which of these chemical reactions could occur, thus helping you identify the solution in the bottle? (a) Barium sulfate could precipitate. (b) Silver chloride could precipitate. (c) Silver sulfate could precipitate. (d) More than one, but not all, of the reactions described in answers a-c could occur. (e) All three reactions described in answers a-c could occur.

We have learned in this chapter that many ionic solids dissolve in water as strong electrolytes; that is, as separated ions in solution. Which statement is most correct about this process? (a) Water is a strong acid and therefore is good at dissolving ionic solids. (b) Water is good at solvating ions because the hydrogen and oxygen atoms in water molecules bear partial charges. (c) The hydrogen and oxygen bonds of water are easily broken by ionic solids.

State whether each of the following statements is true or false. Justify your answer in each case. (a) When acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), is dissolved in water, a conducting solution results. (b) When ammonium nitrate, \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), dissolves in water, the solution is weakly conducting and basic in nature.

Uranium hexafluoride, \(\mathrm{UF}_{6},\) is processed to produce fuel for nuclear reactors and nuclear weapons. UF \(_{6}\) can be produced in a two-step reaction. Solid uranium (IV) oxide, \(\mathrm{UO}_{2}\), is first made to react with hydrofluoric acid (HF) solution to form solid UF \(_{4}\) with water as a by-product. \(U F_{4}\) further reacts with fluorine gas to form UF \(_{6}\). (a) Write the balanced molecular equations for the conversion of \(U O_{2}\) into \(U F_{4}\) and the conversion of \(U F_{4}\) to \(U F_{6}\) (b) Which step is an acid-base reaction? (c) Which step is a redox reaction?
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