Problem 71 - Chemistry Matter and Change

Question

To extract gold from its ore, the ore is treated with sodium cyanide solution in the presence of oxygen and water. \begin{equation} 4 \mathrm{Au}(\mathrm{s})+8 \mathrm{NaCN}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(1) \rightarrow \end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq})+4 \mathrm{NaOH}(\mathrm{aq}) \end{equation} \begin{equation} \begin{array}{l}{\text { a. Determine the mass of gold that can be extracted if }} \\ {25.0 \text { g of sodium cyanide is used. }} \\ {\text { b. If the mass of the ore from which the gold was }} \\ {\text { extracted is } 150.0 \mathrm{g}, \text { what percentage of the ore is gold? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

About 50.24 g of gold can be extracted and the ore is 33.49% gold.

Step 1: Write the balanced chemical equation

The balanced chemical equation for the extraction of gold is given: \[ 4 \mathrm{Au}(\mathrm{s}) + 8 \mathrm{NaCN}(\mathrm{aq}) + \mathrm{O}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(1) \rightarrow 4 \mathrm{NaAu}(\mathrm{CN})_{2}(\mathrm{aq}) + 4 \mathrm{NaOH}(\mathrm{aq}) \] This equation will be used to find the relationship between the reactants and products in terms of moles.

Step 2: Calculate the moles of sodium cyanide

The molar mass of \(\mathrm{NaCN}\) is calculated as follows: \( \mathrm{Na} \) (22.99 g/mol) + \( \mathrm{C} \) (12.01 g/mol) + \( \mathrm{N} \) (14.01 g/mol) = 49.01 g/mol. Given mass of \(\mathrm{NaCN}\) is 25.0 g. Thus, moles of \(\mathrm{NaCN}\) = \( \frac{25.0 \text{ g}}{49.01 \text{ g/mol}} \approx 0.510 \text{ moles} \).

Step 3: Determine moles of gold produced

From the balanced equation, 8 moles of \( \mathrm{NaCN} \) produce 4 moles of \( \mathrm{Au} \). Therefore, the moles of \( \mathrm{Au} \) can be calculated as: \( \frac{4}{8} \times 0.510 = 0.255 \text{ moles of Au} \).

Step 4: Calculate mass of gold produced

The molar mass of \( \mathrm{Au} \) is 196.97 g/mol. Therefore, the mass of gold extracted is: \( 0.255 \times 196.97 \approx 50.24 \text{ g of Au} \).

Step 5: Calculate the percentage of gold in the ore

Given that 50.24 g of gold is extracted from 150.0 g of ore, the percentage of gold in the ore is: \( \frac{50.24}{150.0} \times 100\% \approx 33.49\% \).

Key Concepts

Chemical Reactions

Chemical reactions are the processes in which substances, called reactants, are transformed into different substances, known as products. In the gold extraction process, we start with gold itself (Au), sodium cyanide (NaCN), oxygen (O₂), and water (H₂O) as reactants.

The balanced chemical equation provided:

• Ensures the same number of each type of atom is present on both sides of the equation, upholding the law of conservation of mass.
• Helps chemists understand how reactants interact and what products are formed.

Understanding this specific reaction is key to know how sodium cyanide helps dissolve gold, making it easier to extract. The reaction yields sodium gold cyanide (NaAu(CN)₂) and sodium hydroxide (NaOH) as products, which are in aqueous solutions, ready for further processing to ultimately recover solid gold.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It helps us understand the quantitative relationships between the amounts of substances involved.

For the extraction of gold, stoichiometry allows us to determine how much gold can be produced from a specific amount of cyanide used. This comes from the balanced equation:

• 4 Au + 8 NaCN + O₂ + 2 H₂O → 4 NaAu(CN)₂ + 4 NaOH
• Which shows that 8 moles of sodium cyanide are necessary to produce 4 moles of gold.

By looking at the ratio in the equation, you can predict how changing amounts of one reactant will affect the amount of product produced. Here, it shows the exact relationship that is used to convert moles of NaCN to moles of Au in the extraction process.

Molar Mass Calculations

Molar mass is a fundamental aspect for calculating quantities in chemical reactions. It refers to the mass of one mole of a substance.

In this exercise, molar mass is used in several calculations:

• The molar mass of sodium cyanide (NaCN) is determined by adding the atomic masses of sodium (22.99 g/mol), carbon (12.01 g/mol), and nitrogen (14.01 g/mol), resulting in 49.01 g/mol.
• By dividing the mass of the sample by its molar mass, we find the number of moles for NaCN: \[\text{moles of NaCN} = \frac{25.0 \text{ g}}{49.01 \text{ g/mol}} \approx 0.510 \text{ moles} \]
• Finding the molar mass of gold is also crucial: 196.97 g/mol, so that the mass of gold extracted can be calculated from the moles of gold produced.

Knowing how to calculate molar mass allows you to convert between mass and moles, which is essential in understanding how much of each reactant is required and how much product can be formed.

Ore Percentage

Ore percentage calculation is a real-world application of stoichiometry and molar mass, particularly important in industries like mining. It helps determine the economic feasibility of extracting metal from a given amount of ore.

In this exercise, we calculate what percentage of the ore is actually gold:

• First, determine the mass of gold extracted: 50.24 g.
• The original sample of ore weighed 150.0 g.
• The percentage of gold in the ore is calculated using:\[\text{Percentage of gold} = \left( \frac{50.24}{150.0} \right) \times 100\% \approx 33.49\%\]

This calculation is crucial for determining whether the quantity of gold in the ore justifies the cost and effort of extraction, influencing mining operations and investment decisions.