Question

For many years, the extraction of gold - that is, the separation of gold from other materials- involved the use of potassium cyanide: \(4 \mathrm{Au}+8 \mathrm{KCN}+\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{KAu}(\mathrm{CN})_{2}+4 \mathrm{KOH}\) What is the minimum amount of KCN in moles needed to extract \(29.0 \mathrm{~g}\) (about an ounce) of gold?

Short answer

You need 0.294 moles of KCN to extract 29.0 g of gold.

Step-by-step solution

Calculate Moles of Gold

To determine the moles of gold (Au), we start by finding its molar mass. The molar mass of gold is approximately 197 g/mol. Therefore, the number of moles of gold in 29.0 g is calculated using the formula:\[\text{moles of Au} = \frac{\text{mass of Au}}{\text{molar mass of Au}} = \frac{29.0 \text{ g}}{197 \text{ g/mol}} \approx 0.147 \text{ mol Au}\]

Determine the Stoichiometric Ratio

According to the balanced chemical equation provided:\[4 \mathrm{Au} + 8 \mathrm{KCN} + \mathrm{O}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{KAu} (\mathrm{CN})_{2} + 4 \mathrm{KOH}\]We see that 4 moles of Au require 8 moles of KCN. Therefore, the stoichiometric ratio is 4 Au : 8 KCN, which simplifies to 1 Au : 2 KCN.

Calculate Moles of KCN Required

Using the stoichiometric ratio from Step 2, we know that for every mole of Au, we need 2 moles of KCN. Thus, the moles of KCN required can be calculated as:\[\text{moles of KCN} = 0.147 \text{ mol Au} \times \frac{2 \text{ mol KCN}}{1 \text{ mol Au}} = 0.294 \text{ mol KCN}\]

Key concepts

Molar Mass

Molar mass is an essential concept in chemistry that helps us understand how to relate masses of substances to the number of particles they contain.
A substance's molar mass is its mass per mole, usually measured in grams per mole (g/mol). This value is particularly useful when converting between the mass of a sample and the number of moles, a fundamental step in stoichiometry problems.
For example, gold has a molar mass of approximately 197 g/mol. This means 1 mole of gold weighs about 197 grams.

• To find out how many moles are in a sample, you'd divide the mass of the sample by the molar mass of the substance.
• This conversion allows you to work through chemical reactions quantitatively.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into other substances, called products.
During a chemical reaction, the atoms in the reactants rearrange to create the products, obeying the law of conservation of mass. This law indicates that the number of each type of atom remains constant throughout the reaction.
In the extraction of gold exercise, a chemical reaction occurs where gold and other reactants are converted into gold compounds and potassium hydroxide:

• The equation represents the actual process of extraction, illustrating the importance of understanding chemical changes.
• Learning how these reactions work helps predict the outcomes of chemical processes.
• This understanding is a core aspect of chemistry that applies in many practical and industrial areas.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions where the reactants and products are expressed as chemical formulas.
These equations are balanced to ensure that they adhere to the principle of mass conservation. Every chemical equation must have the same number of each type of atom on both sides of the equation.
The exercise gives us the equation: \[4 \mathrm{Au}+8 \mathrm{KCN}+\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{KAu}(\mathrm{CN})_{2}+4 \mathrm{KOH}\]

• This balanced equation helps determine the proportions or stoichiometric ratios needed for the reaction to occur.
• These ratios allow us to quantify the reactants, such as using moles, to predict the amount needed to conduct the reaction without any waste.

Understanding chemical equations is vital as they provide the quantitative relationships needed to compute amounts of reactants and products in chemical reactions. This is a cornerstone of stoichiometric calculations.