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

Short Answer

Expert verified
The minimum amount of KCN in moles needed to extract 29.0 g of gold is equal to twice the number of moles of Au.

Step by step solution

01

Convert mass of Gold to moles

Use the molar mass of Au (Gold) which is approximately 197.0 g/mol to convert the given 29.0 g of Au to moles. The conversion is based on the formula: \[ \text{number of moles (Au)} = \frac{\text{mass of Au}}{\text{molar mass of Au}} = \frac{29.0 g}{197.0 g/mol} \]
02

Calculate moles of KCN

From the balanced chemical equation, it is known that the molar ratio of Au to KCN is 1:2. Therefore, the number of moles of KCN required can be calculated by multiplying the number of moles of Au by 2. \[ \text{number of moles (KCN)} = \text{molar ratio (Au:KCN)} * \text{number of moles (Au)} = 2 * \text{number of moles (Au)} \]
03

Substitute and solve

Substitute the number of moles of Au from Step 1 into the equation in Step 2 and solve for the number of moles of KCN. That gives the minimum amount of KCN in moles needed to extract 29.0 g of gold.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Calculation in Gold Extraction
Mole calculation is essential in understanding chemical processes, like gold extraction, where precise amounts of reactants are required. In our exercise, we begin with the basics: converting grams of a substance to moles using the formula \( \text{number of moles} = \frac{\text{mass of substance}}{\text{molar mass of substance}} \). For instance, with gold (Au), we calculate the moles from its mass by knowing that the molar mass of Au is 197.0 g/mol. This initial step is crucial for determining how much of the other reactants are needed to complete the reaction.

It’s like following a recipe; we must measure our ingredients properly to ensure the dish (or reaction, in our case) proceeds correctly. Once the number of moles of gold is calculated from its given mass, we can move on to figure out the required proportion of potassium cyanide (KCN) needed for the gold extraction process.
Stoichiometry of Gold Extraction
Stoichiometry is the calculation of reactants and products in chemical reactions. It is a form of bookkeeping for chemists to make sure that atoms are conserved in a reaction. In the gold extraction process, stoichiometry is used to ensure that there is a sufficient amount of KCN to react with all of the gold present.

Using the balanced chemical equation, we can see the ratio of gold to KCN is 1:2. This stoichiometric relationship means that for every mole of gold, two moles of KCN are required. By multiplying the number of moles of gold by this ratio, we obtain the number of moles of KCN needed. This step embodies the core principle of stoichiometry: using ratios from the balanced equation to find out the amounts of other reactants or products.
Chemical Reaction Balancing in Gold Extraction
Chemical reaction balancing is like solving a puzzle. Each piece—or atom—in the reactants must find its place in the products. In the context of gold extraction, we have a balanced chemical equation that underscores the reactants turning into products. For the equation \(4 \text{Au} + 8 \text{KCN} + \text{O}_2 + 2 \text{H}_2\text{O} \longrightarrow 4 \text{KAu}(\text{CN})_2 + 4 \text{KOH}\), the balance is achieved when there are equal numbers of each type of atom on both sides.

This becomes exceedingly useful in stoichiometry, as it allows us to accurately predict how much reactant is needed. The ratio of 1:2 for Au:KCN obtained from the balanced equation informs us precisely how much KCN must be used per mole of gold. Without balancing the equation, we wouldn’t be able to predict this ratio and the chemical reaction would not proceed as intended, much like putting a wrong piece in our puzzle.
Molar Mass in Chemical Reactions
Every chemical substance has a characteristic molar mass, which is the weight of one mole of that substance. In the case of gold (Au), the molar mass is 197.0 g/mol. The concept of molar mass is invaluable in the conversion process of mass to moles (and vice versa) and is the cornerstone of mole calculations.

In our exercise, knowing the molar mass of gold is pivotal as it bridged the mass of the gold we started with (29.0 grams) to the moles needed for further calculations. This conversion is the first step in our journey to understand how much of other chemicals, such as KCN, are needed to react with this gold. It's like knowing the exchange rate between two currencies before trading them. Without molar mass, comparisons and conversions between the mass of substances and the number of particles present would be virtually impossible.

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Most popular questions from this chapter

A major industrial use of hydrochloric acid is in metal pickling. This process involves the removal of metal oxide layers from metal surfaces to prepare them for coating. (a) Write an equation between iron(III) oxide, which represents the rust layer over iron, and \(\mathrm{HCl}\) to form iron(III) chloride and water. (b) If 1.22 moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(289.2 \mathrm{~g}\) of HCl react, how many grams of \(\mathrm{FeCl}_{3}\) will be produced?

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One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is$$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2}$$Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of \(\mathrm{Fe}\) are obtained from a \(2.62 \times 10^{3} \mathrm{~kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

The formula of a hydrate of barium chloride is \(\mathrm{BaCl}_{2} \cdot x \mathrm{H}_{2} \mathrm{O} .\) If \(1.936 \mathrm{~g}\) of the compound gives\(1.864 \mathrm{~g}\) of anhydrous \(\mathrm{BaSO}_{4}\) upon treatment with sulfuric acid, calculate the value of \(x\).

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