Question

Gold is isolated from rocks by reaction with aqueous cyanide, \(\mathrm{CN} : 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) .\) (a) \(\mathrm{Which}\) atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced? (b) The \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) ion can be converted back to \(\mathrm{Au}(0)\) by reaction with \(\mathrm{Zn}(s)\) powder. Write a balanced chemical equation for this reaction. (c) How many liters of a 0.200\(M\) sodium cyanide solution would be needed to react with 40.0 \(\mathrm{kg}\) of rocks that contain 2.00\(\%\) by mass of gold?

Short answer

(a) The Au atom in solid gold is being oxidized, and the O atom in molecular oxygen is being reduced. (b) The balanced chemical equation for the reaction of the gold cyanide complex with zinc powder is \(\mathrm{Zn}(s)+2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(a q)\). (c) The volume of the 0.200 M sodium cyanide solution needed to react with 40.0 kg of rocks containing 2.00% by mass of gold is approximately 162.6 liters.

Step-by-step solution

Identify the atoms being oxidized and reduced

In the given reaction, \(4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q)\), we must determine which atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced. The oxidation states of the atoms are given as follows: Au: 0 (solid gold) NaCN: Na (+1), C (-3) and N (-2) (the sum is equal to 0) O2: 0 (diatomic oxygen) H2O: H (+1) and O (-2) (the sum is equal to 0) Na[Au(CN)2]: Na (+1), Au (+1), C (-2), N (-1) (the sum is equal to -1) NaOH: Na (+1), O (-2), and H (+1) (the sum is equal to 0) Comparing the oxidation states in the reactants and products, we see that the Au atom goes from oxidation state 0 to +1, and the O atom goes from oxidation state 0 to -2. This means that Au is being oxidized, and O2 is being reduced. (a) The Au atom in solid gold is being oxidized, and the O atom in molecular oxygen is being reduced.

Write a balanced equation for the reaction with zinc powder

Now, we will write a balanced chemical equation for the conversion of the gold cyanide complex back to gold by reacting with zinc powder. \(\mathrm{Zn}(s)+2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(a q)\) This is a balanced equation for the reaction. (b) The balanced chemical equation for the reaction of the gold cyanide complex with zinc powder is \(\mathrm{Zn}(s)+2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(a q)\).

Calculate the volume of sodium cyanide solution needed

We are given that the rocks have 2.00% by mass of gold and that there are 40.0 kg of these rocks. To calculate the volume of a 0.200 M sodium cyanide solution needed to react with this gold, we first need to find the moles of gold in the rocks: mass of gold in rocks = 40.0 kg * (2.00 / 100) = 0.800 kg Now, we convert the mass of gold to moles using the molar mass of Au (197 g/mol): moles of gold = (0.800 kg * 1000 g/kg) / 197 g/mol ≈ 4.065 moles Using the stoichiometry of the given reaction, we find that 8 moles of NaCN are needed for every mole of gold: moles of NaCN needed = 4.065 moles Au * (8 moles NaCN / 1 mole Au) ≈ 32.520 moles NaCN With a concentration of 0.200 M NaCN solution, we can find the volume of solution needed: volume of NaCN solution = moles of NaCN / concentration volume of NaCN solution = 32.520 / 0.200 ≈ 162.6 L (c) The volume of the 0.200 M sodium cyanide solution needed to react with 40.0 kg of rocks containing 2.00% by mass of gold is approximately 162.6 liters.

Key concepts

Oxidation and Reduction

Oxidation and reduction are fundamental concepts in chemistry that involve the change of oxidation states of atoms. In the context of gold isolation, we focus on understanding how specific atoms lose or gain electrons in a chemical reaction.
When an atom is oxidized, it loses electrons, leading to an increase in its oxidation state. On the other hand, when an atom is reduced, it gains electrons, resulting in a decrease in its oxidation state.

In the reaction for extracting gold from ores using cyanide, the oxidation state of gold (Au) changes from 0 to +1, indicating it is oxidized. This means it loses electrons. Meanwhile, oxygen ( ext{O}_2), which starts with an oxidation state of 0, is reduced to an oxidation state of -2 when it forms ext{NaOH}. This reduction implies oxygen gains electrons.

• The oxidation process: ext{Au} (from 0 to +1)
• The reduction process: ext{O}_2 (from 0 to -2)

Understanding these transformations is crucial because they reveal the electron flow in chemical reactions, which is vital for predicting the outcome of reactions, especially in metallurgical processes like gold isolation.

Chemical Reaction Balancing

Balancing a chemical equation is an essential skill in understanding the stoichiometry of a reaction. It ensures that the number of atoms for each element is the same on both sides of the equation, abiding by the law of conservation of mass.

In our example, balancing the reaction between the gold cyanide complex and zinc lets us track all reactants and products.
Initially, we look at each component of the reaction:

• Zinc (Zn) reacts with 2 ext{Au(CN)_2}^{-} ions.
• This process yields 2 Au atoms and ext{Zn(CN)_4}^{2-}.

By systematically verifying that there are equal numbers of atoms for each element across reactants and products, we confirm the balanced equation and ensure it accurately represents the chemical change.

For instance, the reaction \[ ext{Zn}(s) + 2 ext{Au(CN)_2}^{-} (aq) \rightarrow 2 ext{Au}(s) + ext{Zn(CN)_4}^{2-} (aq)\]illustrates a balanced equation where zinc facilitates the reduction of the gold compound back to elemental gold.

Mole Calculations

Mole calculations are a cornerstone in chemistry, allowing us to convert between mass, number of particles, and volume of solutions using the mole concept. In processes like the isolation of gold, these calculations help us determine the quantities of reactants needed.

Given rocks with a 2% gold composition, understanding how to calculate moles from a given mass is crucial. Here’s how the process works:

• Convert the mass percentage into an actual quantity of gold: from 40 kg of rock, 0.8 kg is gold.
• Calculate moles of gold using its molar mass (197 g/mol): \[ moles = \frac{mass\,(g)}{molar\,mass\,(g/mol)} \approx 4.065 \text{ moles of Au} \]
• With the stoichiometric ratio from the balanced equation (8 moles of ext{NaCN} for every mole of gold), calculate the needed moles of ext{NaCN}.
• Finally, determine the volume of sodium cyanide solution (0.200 M) required. Using \[ volume\,(L) = \frac{moles\,of\,NaCN}{concentration\,(M)} \approx 162.6 \text{ L} \]

This shows how mole calculations can guide the efficient use of reagents in industrial processes.