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Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous \(\mathrm{NO}\). Write a balanced equation for the dissolution of gold in aqua regia.

Short Answer

Expert verified
The balanced chemical equation for the dissolution of gold in aqua regia is \(Au(s) + HNO_3(aq) + 4HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g) + 3H_2O(l)\).

Step by step solution

01

Write down the reactants and products

First, we write down the reactants and products of the reaction: 1. Reactants: Gold (Au), Nitric acid (HNO3), and Hydrochloric acid (HCl) 2. Products: \(\mathrm{AuCl}_{4}^{-}\) and gaseous \(\mathrm{NO}\) Now, we need to write the chemical equation. \(Au(s) + HNO_3(aq) + HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g)\)
02

Balance the chemical equation

\(Au(s) + HNO_3(aq) + HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g)\) Balancing the above chemical equation: 1. Balance the gold (Au) atoms: There is 1 gold atom in both the reactants and the products, so this part is already balanced. 2. Balance the nitrogen (N) atoms: There is 1 nitrogen atom in HNO3 and 1 nitrogen atom in NO, so this part is also already balanced. 3. Balance the hydrogen (H) atoms: There are 3 hydrogen atoms in the reactants (1 in HNO3 and 2 in HCl) and only one hydrogen atom in the product (AuCl4-). So, to balance the hydrogen atoms, we need to add 3 H2O molecules as products: \(Au(s) + HNO_3(aq) + HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g) + 3H_2O(l)\) 4. Balance the chlorine (Cl) atoms: There are 1 chlorine atom in HCl and 4 chlorine atoms in AuCl4-. To balance the chlorine atoms, we need to multiply HCl by 4 in the reactants: \(Au(s) + HNO_3(aq) + 4HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g) + 3H_2O(l)\) 5. Balance the oxygen atoms: After balancing the other elements, check the oxygen (O) atoms are balanced. There are 4 oxygen atoms in the reactants and 4 in the products, so this part is also already balanced. The final balanced chemical equation is: \(Au(s) + HNO_3(aq) + 4HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g) + 3H_2O(l)\)

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Key Concepts

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

Balanced Chemical Equation
A balanced chemical equation is fundamental for understanding how chemical reactions occur. It ensures that the law of conservation of mass is upheld, meaning that atoms are neither created nor destroyed in the reaction. When gold dissolves in aqua regia, we're dealing with a complex chemical process that can be daunting at first glance. However, by breaking it down step by step, we can create a clear picture of the reaction.
To illustrate this, let's revisit the provided exercise, where the final balanced equation is presented as:
\[Au(s) + HNO_3(aq) + 4HCl(aq) \rightarrow AuCl_{4}^{-}(aq) + NO(g) + 3H_2O(l)\]
This equation represents the stoichiometry of the dissolution of gold in aqua regia. Notice how every atom on the left side of the reaction is accounted for on the right side.To facilitate a deeper understanding, we need to emphasize the significance of coefficients in front of the chemical formulas. For instance, the '4' in front of HCl indicates that four hydrochloric acid molecules react with each nitric acid molecule and the gold atom to produce one tetra-chloroaurate ion, nitric oxide, and three water molecules.
We should also explain why it's necessary to add water molecules to the products. As hydrogen atoms are less in the product, we add water to balance the hydrogen and oxygen atoms. This illustrates an important aspect of balancing reactions where new compounds like water can be formed to satisfy the conservation of mass.
Reactivity of Gold
Gold is known for its remarkable stability and resistance to corrosion and oxidation. This trait, known as noble metal behavior, arises from gold's reluctance to react with most substances. This inertness makes gold a precious metal, both literally and chemically.
In the context of the exercise, it's fascinating to note that gold will not dissolve in either concentrated nitric acid or hydrochloric acid when each is used alone. This is due to gold's low reactivity, which manifests in its ability to withstand the individual effects of these strong acids.
However, the combination of the two acids, known as aqua regia, creates a potent mixture capable of dissolving gold. Aqua regia is a mixture of nitric acid and hydrochloric acid, typically in a volumetric ratio of 1:3. The nitric acid serves as an oxidizing agent, producing chloride ions from hydrochloric acid, which are highly effective in dissolving gold. Moreover, aqua regia is one of the few substances that can dissolve gold and platinum, highlighting the exceptional nature of this reaction. Understanding the unique chemical environment that aqua regia provides helps students grasp why gold's reactivity changes so dramatically in this mixture.
Acid-Metal Reaction
Acid-metal reactions are a type of chemical reaction where metals react with acids to produce a salt and hydrogen gas. However, gold showcases its unique properties by not following this usual pattern.
Let's consider the typical reaction: a metal, such as zinc, will react vigorously with an acid like hydrochloric acid to form zinc chloride and hydrogen gas. This reaction is straightforward and demonstrates a metal's capacity to displace hydrogen from an acid, forming a salt.
In our exercise with gold, however, the outcome is quite different. Instead of releasing hydrogen gas, gold reacts with both components of aqua regia to form chloroaurate ions and nitric oxide. Here, the acid does not act merely as a proton donor; instead, the reaction involves a complex series of steps catalyzed by the presence of both acids. It's essential for students to recognize that while gold is immune to individual acids, it succumbs to the combined chemical forces of aqua regia, underscoring the need to analyze reactions involving gold, and aqua regia not as typical acid-metal reactions but as special cases warranting closer inspection.

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Most popular questions from this chapter

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{8}^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{C}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\)

A silver concentration cell is set up at \(25^{\circ} \mathrm{C}\) as shown below:The \(\mathrm{AgCl}(s)\) is in excess in the left compartment. a. Label the anode and cathode, and describe the direction of the electron flow. b. Determine the value of \(K_{\text {sp }}\) for \(\mathrm{AgCl}\) at \(25^{\circ} \mathrm{C}\).

Gold is produced electrochemically from an aqueous solution of \(\mathrm{Au}(\mathrm{CN})_{2}^{-}\) containing an excess of \(\mathrm{CN}^{-}\). Gold metal and oxygen gas are produced at the electrodes. What amount (moles) of \(\mathrm{O}_{2}\) will be produced during the production of \(1.00 \mathrm{~mole}\) of gold?

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of \(2.50 \mathrm{~A}\) in \(15.0 \mathrm{~min} ?\)

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