Problem 117 - Chemistry and Chemical Reactivity

Question

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 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?

Step-by-Step Solution

Answer

(a) Oxygen is the oxidizing agent; gold is the reducing agent. (b) You need 25.707 liters of 0.075 M NaCN solution.

Step 1: Identify the Oxidizing and Reducing Agents

In a chemical reaction, the oxidizing agent is the species that gets reduced by gaining electrons, and the reducing agent is the species that gets oxidized by losing electrons. In this reaction, oxygen ( O_2 gas) is reduced as it facilitates the oxidation of gold ( Au ) from its neutral to its positive oxidation state, making gold be the reducing agent.

Step 2: Determine What Has Been Oxidized and Reduced

The oxidation state of gold ( Au ) changes from 0 in the solid gold to +1 in NaAu(CN)_2 . Therefore, gold has been oxidized. Oxygen ( O_2 ) is reduced as it accepts electrons in the formation of NaOH .

Step 3: Calculate Moles of Gold in Rock

First, find the mass of gold in the gold-bearing rock. Since the rock is 0.019% gold, this means there are \( \frac{0.019}{100} \times 1000 \text{ kg} \) of gold, which equals 0.19 kg. Then convert kilograms of gold to grams: \( 0.19 \text{ kg} \times 1000 \text{ g/kg} = 190 \text{ g} \) of gold.

Step 4: Convert Mass of Gold to Moles of Gold

Find the molar mass of gold (\( \text{Au} \)), which is approximately 197 g/mol. The moles of gold can be calculated as: \( \text{Moles of Au} = \frac{190 \text{ g}}{197 \text{ g/mol}} \approx 0.964 \text{ moles} \).

Step 5: Use Stoichiometry to Determine Moles of \( \text{NaCN}\) Needed

According to the balanced equation, 8 moles of \( \text{NaCN} \) are needed to react with 4 moles of gold. Therefore, you need \( 2 \times 0.964 = 1.928 \) moles of \( \text{NaCN} \).

Step 6: Calculate Volume of NaCN Solution Required

The concentration of \( \text{NaCN} \) solution is given as 0.075 M. Use the formula \( \text{Moles} = \text{Molarity} \times \text{Volume} \) to find the volume: \( \text{Volume} = \frac{1.928 \text{ moles}}{0.075 \text{ mol/L}} = 25.707 \text{ L}\).

Key Concepts

Oxidizing Agents

In any chemical reaction, oxidizing agents play a pivotal role because they accept electrons from other substances. In the gold extraction reaction, our oxidizing agent is oxygen gas (O_2). By gaining electrons, the oxygen helps facilitate the oxidation process, which is crucial for getting gold into a soluble form. Oxygen effectively "pulls" electrons away from the gold, enabling the gold atoms to bond with cyanide ions in the solution.

• Key Insight: The oxidizing agent always gets reduced because it gains electrons.
• Example here: in our case, oxygen ( O_2) is transformed as it assists in oxidizing gold ( Au).

Understanding oxidizing agents is essential, especially in reactions like this, where valuable metals are extracted from ores.

Reducing Agents

Reducing agents are known for their ability to donate electrons. They undergo oxidation themselves. Gold ( Au) serves as the reducing agent in the given reaction. By losing electrons, gold goes from a neutral state in solid form to a positively charged ion when dissolved in the solution.

• Main Function: A reducing agent facilitates the reduction of another species.
• In this reaction: Gold loses electrons, transforming from its metallic state to bind with cyanide ions effectively.

Grasping the role of reducing agents is crucial for understanding how metals can be converted to different states during chemical changes.

Stoichiometry

Stoichiometry is the mathematical core of understanding chemical equations. It involves using the coefficients from the balanced reaction to determine relations between reactants and products. In our gold extraction problem, stoichiometry is necessary to calculate how much sodium cyanide ( NaCN) is needed to dissolve a specific mass of gold.

• Balanced Equation: 4 moles of gold need 8 moles of NaCN.
• Calculation: If you have 0.964 moles of gold, according to stoichiometry, you'll require twice as much NaCN, or 1.928 moles.

Accurate stoichiometric calculations are vital for ensuring that the correct amounts of chemical inputs are used for the desired results.

Molarity Concentration

Molarity quantifies how concentrated a solution is in terms of moles per liter ( M). It becomes very useful in determining the precise amount of reactants needed in a reaction. To extract gold using NaCN, we need to calculate the volume of NaCN solution based on its molarity.

• Understanding Formula: Moles = Molarity × Volume. Rearranging gives us Volume = Moles / Molarity.
• Application: With a molarity of 0.075 M and needing 1.928 moles, we calculate the necessary volume as approximately 25.707 liters.

Understanding molarity helps ensure precise chemical reactions, economizing resources while achieving efficient extraction processes.