Gold can be dissolved from gold-bearing rock by treating the...

Gold can be dissolved from gold-bearing rock by treating the rock with sodium cyanide in the presence of oxygen. $4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$ $$ 4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq}) $$ (a) Name the oxidizing and reducing agents in this reaction. What has been oxidized, and what has been reduced? (b) If you have exactly one metric ton $(1 \text { metric ton }=$ $1000 \mathrm{kg})$ of gold-bearing rock, what volume of $0.075 \mathrm{M} \mathrm{NaCN},$ in liters, do you need to extract the gold if the rock is $0.019 \%$ gold?

Gold can be dissolved from gold-bearing rock by treating the rock with sodium cyanide in the presence of oxygen.
$4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$
$$
4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq})
$$
(a) Name the oxidizing and reducing agents in this reaction. What has been oxidized, and what has been reduced?
(b) If you have exactly one metric ton $(1 \text { metric ton }=$ $1000 \mathrm{kg})$ of gold-bearing rock, what volume of $0.075 \mathrm{M} \mathrm{NaCN},$ in liters, do you need to extract the gold if the rock is $0.019 \%$ gold?

Key Concepts

Molarity and Solution Concentration

Molarity is a measure of the concentration of a solute in a solution, expressed as moles per liter. Understanding molarity allows chemists to relate volumes of a solution to the number of moles of a reactant available for a chemical reaction. This concept is pivotal for determining the volume of a reagent solution, such as sodium cyanide, needed to achieve the desired reaction with the available gold in the ore.

Redox Reactions

Redox (oxidation–reduction) reactions involve the transfer of electrons between species. One species loses electrons (is oxidized) and another gains electrons (is reduced). This fundamental concept underlies many chemical processes, including the extraction of metals, and is crucial for understanding how reactants transform during a reaction.

Oxidizing and Reducing Agents

An oxidizing agent is a substance that accepts electrons and is themselves reduced, while a reducing agent donates electrons and is oxidized. Identifying which substances undergo changes in oxidation state allows us to determine these agents. In processes such as metal extraction, recognizing these roles helps in understanding how the desired metal is mobilized or precipitated through redox reactions.

Stoichiometry in Chemical Reactions

Stoichiometry involves using balanced chemical equations to relate the amounts of reactants and products in a reaction. It provides the quantitative link between reactant consumption and product formation. This concept is essential when calculating the precise quantities of chemicals needed to drive a reaction to completion, such as determining reagent volumes based on the amount of target metal in ore.

Percent Composition Calculations

Percentage composition is the calculation used to determine the mass fraction of a particular component in a mixture or compound. It is particularly important in mining and resource extraction to quantify the amount of a valuable component, such as gold in ore. This information is fundamental when calculating the reagent quantity required for processing a given mass of ore.

Transcript

00:01 In this question we have been given the reaction we have been given the chemical equation that represents the reaction that happens when we dissolve gold into a solution of sodium sinide so basically in the first subpart of this question we need to find out which is being ox what species is being oxidized and what is being reduced right so we can say that oxidation oxidation can be related to an increase in the oxidation states right increase in oxidation state means that this species has lost electrons and that is the only way to increase the oxidation state and that is why we can say that the species has been oxidized similarly for reduction we can say that it can be related to or it can be connected to a decrease in oxidation state of any species right so now in the given in the given reaction we can see that for gold metal for gold metal since it is a free metal so it is having a zero oxidation state and similarly the oxygen gas that is having zero oxidation state because it is only free gas it is not it is not connected to something it is it hasn't formed any kind of bond with any other element now on the product side we can clearly see that this gold is involved in this compound so basically here it is having an oxidation state of plus one now that is an increase of unit from its oxidation state on the reactant side so we can say that yes gold has been oxidized gold has been oxidized and also since the oxygen atoms are having zero oxidation state but in but on the product side all the four oxygen atoms are having a minus two oxidation state which which is a decrease of two units so we can say that o2 has been reduced right these are the two species that have been oxidized and reduced respectively and also the under concept is that oxidizing agent oxidizing agent is the species that oxidizes others but it gets reduced itself it gets reduced itself so and vice versa and similarly we can say that a reducing agent is something that reduces other species but it gets oxidized itself now we can say that since o2 is being reduced it is going to be an oxidizing agent here it is going to be an oxidizing agent and similarly gold is going to work as a reducing agent so with this we have successfully solved our first subpart now moving on to the second subpart which says that we are having thousand kg we are having thousand kilograms…

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