Question

To extract gold from its ore, the ore is treated with sodium cyanide solution in the presence of oxygen and water. \(4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) \rightarrow\) \(4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq})\) a. Determine the mass of gold that can be extracted if 25.0 \(\mathrm{g}\) of sodium cyanide is used. b. If the mass of the ore from which the gold was extracted is \(150.0 \mathrm{g},\) what percentage of the ore is gold?

Short answer

To extract gold using 25 grams of sodium cyanide, 50.2 grams of gold can be obtained. The percentage of gold in the 150 grams of ore is 33.5%.

Step-by-step solution

Interpret the balanced chemical equation

The balanced chemical equation is given as: \[4 \mathrm{Au}_{\large{(s)}}+8 \mathrm{NaCN}_{\large{(aq)}}+\mathrm{O}_{2_{\large{(g)}}}+2 \mathrm{H}_{2\large{(l)}} \rightarrow 4 \mathrm{NaAu}(\mathrm{CN})_{2_{\large{(aq)}}}+4 \mathrm{NaOH}_{\large{(aq)}}\] From the balanced chemical equation, we can gather the stoichiometric coefficients: 4 moles of Au react with 8 moles of NaCN to produce 4 moles of NaAu(CN)₂ and 4 moles of NaOH.

Determine the limiting reactant

It is given that 25 grams of sodium cyanide (NaCN) is used. We need to determine the moles of NaCN present: \[ \text{Moles of NaCN} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.0 \,\text{g}}{49.01 \,\text{g/mol}} = 0.510 \,\text{mol} \] As 8 moles of NaCN are required to react with 4 moles of Au, we can calculate the moles of Au that can be extracted from these 0.510 moles of NaCN: \[ \text{Moles of Au} = \frac{\text{moles of NaCN}}{2} = \frac{0.510 \,\text{mol}}{2} = 0.255 \,\text{mol} \]

Calculate the mass of gold that can be extracted

Now that we have the moles of Au, we can calculate the corresponding mass: \[ \text{Mass of Au} = \text{moles} \times \text{molar mass} = 0.255 \,\text{mol} \times 197.0 \,\text{g/mol} = 50.2 \,\text{g} \] So, 50.2 grams of gold can be extracted using 25 grams of sodium cyanide.

Calculate the percentage of gold in the ore

It is given that the mass of the ore is 150 grams. To find the percentage of gold in the ore, we can use the following formula: \[ \text{Percentage of gold} = \frac{\text{mass of gold}}{\text{mass of ore}} \times 100 \] \[ \text{Percentage of gold} = \frac{50.2 \,\text{g}}{150.0 \,\text{g}} \times 100 = 33.5 \% \] Therefore, the percentage of gold in the ore is 33.5%.

Key concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that helps us understand the quantitative relationships between reactants and products in a chemical reaction. In the context of gold extraction, stoichiometry allows us to predict how much gold can be obtained from a given amount of sodium cyanide (NaCN).

The balanced chemical equation provided is:

• 4 Au(s) + 8 NaCN(aq) + O\(_2\)(g) + 2 H\(_2\)O(l) \( \rightarrow \) 4 NaAu(CN)\(_2\)(aq) + 4 NaOH(aq)

From this equation, we can see that 4 moles of sodium cyanide are required for each mole of gold. This relationship is crucial for the stoichiometric calculations necessary to determine the expected amount of gold produced.

Stoichiometry effectively relates the molar ratios of reactants to the products that form, which is directly used in the calculation of limiting reactants and the mass of the products.

Limiting Reactant

The concept of a limiting reactant is critical in determining how much product can be formed in a chemical reaction. The limiting reactant is the substance that runs out first, and thus limits the extent of the reaction. In our case of gold extraction, sodium cyanide is identified as the limiting reactant.

Here's why:

• We calculate the amount of sodium cyanide used: with 25 grams, we find we have 0.510 moles.
• The reaction demands 8 moles of NaCN for every 4 moles of gold, implying that for every 0.510 moles of NaCN, only 0.255 moles of gold could be processed.

Once the sodium cyanide is consumed, no more gold can be extracted, even if additional gold from the ore is available. Understanding limiting reactants enables chemists to maximize the efficiency of chemical processes by predicting when a reaction will stop based on the quantity of the least abundant reactant.

Chemical Reaction

A chemical reaction is a process where substances (reactants) are converted into different substances (products). In the gold extraction process, the chemical reaction of interest involves converting solid gold (Au) into a soluble compound using sodium cyanide (NaCN), oxygen (O\(_2\)), and water (H\(_2\)O).

This is an example of a redox reaction with the balanced chemical equation:

• 4 Au(s) + 8 NaCN(aq) + O\(_2\)(g) + 2 H\(_2\)O(l) \( \rightarrow \) 4 NaAu(CN)\(_2\)(aq) + 4 NaOH(aq)

In this balanced reaction, gold (Au) reacts with sodium cyanide (NaCN) in the presence of oxygen and water to form sodium dicyanoaurate (NaAu(CN)\(_2\)) and sodium hydroxide (NaOH).

Complete atom reorganization happens during such a reaction, and no atoms are lost; just reshuffled to produce different compounds. The balanced equation guarantees that mass and charge are preserved, crucial for yielding accurate quantitative reactions in industrial processes such as gold extraction.