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. $$\begin{aligned}4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) & \rightarrow \\4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq}) \end{aligned}$$ (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 metric ton $=1000 \mathrm{kg}$ of gold-bearing rock, what volume of 0.075 M 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.
$$\begin{aligned}4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) & \rightarrow \\4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq})
\end{aligned}$$
(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 metric ton $=1000 \mathrm{kg}$ of gold-bearing rock, what volume of 0.075 M NaCN, in liters, do you need to extract the gold if the rock is 0.019\% gold?

Key Concepts

Redox Reactions

Redox reactions involve the transfer of electrons from one species to another, leading to changes in oxidation states. Understanding these reactions is essential as they explain how substances are oxidized (lose electrons) and reduced (gain electrons) in chemical processes.

Oxidizing and Reducing Agents

In any redox reaction, the oxidizing agent is the species that accepts electrons (is reduced) and facilitates the oxidation of another substance, while the reducing agent donates electrons (is oxidized) and causes another species to be reduced. This concept is key to identifying the roles different reactants play in the overall reaction.

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction and relies on a balanced chemical equation. It is essential for determining the exact mole ratios needed to convert reactants into products, which is vital for both reaction analysis and practical applications such as reagent calculation.

Concentration and Solution Volume Calculations

Molarity, defined as moles of solute per liter of solution, is a fundamental concept for calculating the volume of a reagent required to achieve a desired reaction based on the amount of reactant. This is especially important in practical extraction and synthesis processes, where precise measurements are required.

Percentage Composition

Percentage composition is a method used to indicate the proportion of a particular substance within a mixture or compound, usually expressed as a percentage. This is crucial when dealing with ores or mixtures, as it helps determine the actual amount of the desired element present relative to the total mass, enabling accurate stoichiometric calculations.

Transcript

00:01 In this problem, we are going to be considering an assay, that's an analysis of composition, of gold.

00:09 We were given a chemical equation that shows the reaction for gold being taken from a gold -bearing rock by treating the rock with sodium, cyanide, and the presence of oxygen.

00:24 So we've got solid gold being reacted with aqueosodium, cyanide, and oxygen and water.

00:31 The products are n -a -u -c -n -2 that's a gold complex and 4 n -a -o -h for this problem we're going to name the oxidizing and reducing agents and we're going to determine what has been oxidized and what has been reduced remember the oxidizing agent is the substance that is undergoing reduction oxidizing agent.

01:21 So the oxidizing agent is the substance that's being reduced.

01:25 And the reducing agent is a substance that is being oxidized.

01:38 We have a second, okay, gold is being oxidized.

01:46 Metals are always oxidized.

01:50 And our oxygen is also changing oxidation states.

01:55 So gold is being oxidized, therefore it's a reducing agent.

01:59 Oxygen is being reduced, therefore it's the oxidizing.

02:02 Agent.

02:04 Our second order of business is as follows.

02:09 We have one.

02:11 We're going to find the percent.

02:15 Let me see here.

02:16 If we have one metric ton of gold -bearing rock, we're going to find the volume of 0 .075 molar sodium cyanide in liters we need to extract the gold if the rock is 0 .019 % gold.

02:38 Not that hard.

02:49 So the mass of the ore is going to equal our metric ton, which is 1 ,000 kilograms, which is 1 times 10 to the 6th grams.

03:09 That's how much ore we have.

03:16 The ore is 0 .019 % gold, so we can multiply 1 .1 times 10 to the 6 times the 0 .0019.

03:33 And our sodium cyanide has a molarity of 0 .075 molar.

03:44 Now we're going to take the mass of our ore...

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