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Aqueous solutions of ammonium sulfide, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S},\) and \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) react to produce HgS and \(\mathrm{NH}_{4} \mathrm{NO}_{3}\). (a) Write the overall balanced equation for the reaction. Indicate the state (s, aq) for each compound. (b) Name each compound. (c) What type of reaction is this?

Short Answer

Expert verified
(a) \((\mathrm{NH}_4)_2\mathrm{S}(aq) + \mathrm{Hg}(\mathrm{NO}_3)_2(aq) \rightarrow \mathrm{HgS}(s) + 2\mathrm{NH}_4\mathrm{NO}_3(aq)\). (b) Ammonium sulfide, mercury(II) nitrate, mercury(II) sulfide, ammonium nitrate. (c) Double displacement reaction.

Step by step solution

01

Identify Reactants and Products

Identify the chemical formulas for the reactants: ammonium sulfide is \((\mathrm{NH}_4)_2\mathrm{S}\) and mercury(II) nitrate is \(\mathrm{Hg}(\mathrm{NO}_3)_2\). The products of the reaction are mercury(II) sulfide, \(\mathrm{HgS}\), and ammonium nitrate, \(\mathrm{NH}_4\mathrm{NO}_3\).
02

Write the Unbalanced Equation

Write the chemical equation using the formulas identified: \((\mathrm{NH}_4)_2\mathrm{S}(aq) + \mathrm{Hg}(\mathrm{NO}_3)_2(aq) \rightarrow \mathrm{HgS}(s) + \mathrm{NH}_4\mathrm{NO}_3(aq)\).
03

Balance the Equation

Balance the equation by ensuring the number of atoms for each element is the same on both sides. The balanced version is: \[(\mathrm{NH}_4)_2\mathrm{S}(aq) + \mathrm{Hg}(\mathrm{NO}_3)_2(aq) \rightarrow \mathrm{HgS}(s) + 2\mathrm{NH}_4\mathrm{NO}_3(aq)\].
04

Name Each Compound

Name the compounds involved:- \((\mathrm{NH}_4)_2\mathrm{S}\) is ammonium sulfide.- \(\mathrm{Hg}(\mathrm{NO}_3)_2\) is mercury(II) nitrate.- \(\mathrm{HgS}\) is mercury(II) sulfide.- \(\mathrm{NH}_4\mathrm{NO}_3\) is ammonium nitrate.
05

Determine the Type of Reaction

This reaction can be classified as a double displacement reaction. In a double displacement reaction, the cations and anions in the reactants swap partners to form new products.
06

Add States of Matter

Indicate the state of each compound as noted. The final balanced equation with states is \((\mathrm{NH}_4)_2\mathrm{S}(aq) + \mathrm{Hg}(\mathrm{NO}_3)_2(aq) \rightarrow \mathrm{HgS}(s) + 2\mathrm{NH}_4\mathrm{NO}_3(aq)\).

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Key Concepts

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

Aqueous Solutions
When we refer to aqueous solutions, we're discussing a substance that is dissolved in water. So, when you see the notation 'aq', it means that the compound is in an aqueous state.
This concept is crucial because many chemical reactions occur in aqueous solutions, especially in the context of double displacement reactions like the one in your exercise.
Aqueous solutions are central to many areas of chemistry since they facilitate the movement and interaction of ions. Water is a universal solvent, and many ionic compounds dissolve readily in it to produce ions that can engage in a variety of chemical reactions.
- Water ( H₂O) acts as a solvent because of its polar nature. - Solutes that dissolve in water form these aqueous solutions. Understanding which compounds can dissolve in water helps predict whether or not a reaction will occur. In the given reaction, ammonium sulfide \((\mathrm{NH}_{4})_{2}\mathrm{S}\) and mercury(II) nitrate \(\mathrm{Hg}(\mathrm{NO}_{3})_{2}\) are both presented as being in aqueous form (aq), signaling their ready availability for engagement in the chemical reaction.
Double Displacement Reaction
A double displacement reaction is a fascinating type of chemical reaction. It occurs when parts of two ionic compounds are exchanged and form two new compounds. These reactions usually occur in aqueous solutions where ions can move freely.
In essence, the cations of the reacting compounds swap anions, resulting in the formation of new products. This type of reaction is evident in your exercise:
  • The ammonium ion \((\mathrm{NH}_4^+)\) from ammonium sulfide combines with the nitrate ion \((\mathrm{NO}_3^-)\) from mercury(II) nitrate to form ammonium nitrate \((\mathrm{NH}_4\mathrm{NO}_3)\).
  • The mercury ion \((\mathrm{Hg}^{2+})\) pairs with the sulfide ion \((\mathrm{S}^{2-})\), forming mercury(II) sulfide \((\mathrm{HgS})\).
What's special about double displacement reactions is that they often result in the formation of a precipitate, like \(\mathrm{HgS}\) here, which is a solid. This precipitation indicates that the reaction has happened.
Compound Naming
Compound naming in chemistry follows specific rules to ensure clarity and uniformity. Let's break down how the compounds in your exercise are named.Firstly, consider ionic compounds, which consist of cations and anions: - Ammonium sulfide \((\mathrm{NH}_{4})_{2}\mathrm{S}\) features the ammonium ion \((\mathrm{NH}_4^+)\) and sulfide ion \((\mathrm{S}^{2-})\). Here, 'ammonium' is the cation name and 'sulfide' is the anion name. - Mercury(II) nitrate \(\mathrm{Hg}(\mathrm{NO}_{3})_{2}\) comprises mercury ions \((\mathrm{Hg}^{2+})\) and nitrate ions \((\mathrm{NO}_3^-)\). The roman numeral in 'Mercury(II)' indicates the +2 charge.Rules for naming:
  • Cations keep their elemental name, but some, like metals with multiple charge states, gain a Roman numeral.
  • Anions take the root of their elemental name and append '-ide,' or if polyatomic, retain names like 'nitrate'.
These conventions allow us to correctly identify the composition of chemical compounds.
Balancing Chemical Equations
Balancing chemical equations is an essential skill in chemistry. It ensures that there is the same number of each type of atom on both sides of the equation, complying with the law of conservation of mass.
To balance a chemical equation:
  • Identify all reactants and products and write their unbalanced chemical equation.
  • Count the number of atoms for each element in the reactants and products.
  • Add coefficients (numbers before compounds) to balance the atoms of each element on both sides.
In the reaction from your exercise: - Start with the unbalanced equation: \((\mathrm{NH}_4)_{2}\mathrm{S}(aq) + \mathrm{Hg}(\mathrm{NO}_3)_{2}(aq) \rightarrow \mathrm{HgS}(s) + \mathrm{NH}_4\mathrm{NO}_3(aq)\).Balance it by adding a coefficient of 2 before \(\mathrm{NH}_4\mathrm{NO}_3\): \((\mathrm{NH}_4)_{2}\mathrm{S}(aq) + \mathrm{Hg}(\mathrm{NO}_3)_{2}(aq) \rightarrow \mathrm{HgS}(s) + 2\mathrm{NH}_4\mathrm{NO}_3(aq)\). The balanced equation ensures all atoms are accounted for, making it accurate and reliable for analysis or experimentation.

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Most popular questions from this chapter

Predict the products of each precipitation reaction. Balance the completed equation, and then write the net ionic equation. (a) \(\mathrm{NiCl}_{2}(\mathrm{aq})+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(\mathrm{aq}) \rightarrow\) (b) \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \rightarrow\)

Equal amounts of two acids- -HCl and HCO \(_{2} \mathrm{H}\) (formic acid) - are placed in solution. When equilibrium has been achieved, the HCl solution has a much greater electrical conductivity than the HCO,H solution. Which reaction is more product-favored at equilibrium? $$\begin{aligned}\mathrm{HCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) & \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\\\\mathrm{HCO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) & \rightleftarrows \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HCO}_{2}^{-}(\mathrm{aq})\end{aligned}$$

Identify, from each list below, the compound or compounds that will dissolve in water to give a solution that is only a very weak conductor of electricity. (a) \(\mathrm{NH}_{3}, \mathrm{NaOH}, \mathrm{Ba}(\mathrm{OH})_{2}, \mathrm{Fe}(\mathrm{OH})_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{Na}_{3} \mathrm{PO}_{4}, \mathrm{HF}, \mathrm{HNO}_{3}\)

The presence of arsenic in a sample that may also contain another Group 5 A element, antimony, can be confirmed by first precipitating the \(\mathrm{As}^{3+}\) and \(\mathrm{Sb}^{3+}\) ions as yellow solid \(\mathrm{As}_{2} \mathrm{S}_{3}\) and orange solid \(\mathrm{Sb}_{2} \mathrm{S}_{3} .\) If aqueous HCl is then added, only \(\mathrm{Sb}_{2} \mathrm{S}_{3}\) dissolves, leaving behind solid \(\mathrm{As}_{2} \mathrm{S}_{3} .\) The \(\mathrm{As}_{2} \mathrm{S}_{3}\) can then be dissolved using aqueous HNO \(_{3}\) $$\begin{aligned} 3 \mathrm{As}_{2} \mathrm{S}_{3}(\mathrm{s})+10 \mathrm{HNO}_{3}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\ell) & \rightarrow \\ 6 \mathrm{H}_{3} \mathrm{AsO}_{4}(\mathrm{aq})+10 \mathrm{NO}(\mathrm{g})+9 \mathrm{S}(\mathrm{s}) \end{aligned}$$ Finally, the presence of arsenic is confirmed by adding \(\mathrm{AgNO}_{3}\) to the solution of \(\mathrm{H}_{3} \mathrm{AsO}_{4}\) to precipitate a reddish brown solid \(\mathrm{Ag}_{x} \mathrm{AsO}_{y} .\) The composition of this solid is As, \(16.199 \%\) and \(\mathrm{Ag}, 69.964 \%\) (a) What are the oxidation numbers of As, \(S,\) and \(N\) in the reaction of \(\mathrm{As}_{2} \mathrm{S}_{3}\) with nitric acid? (b) What is the formula of the reddish brown solid \(\mathrm{Ag}_{\mathrm{x}} \mathrm{A} \mathrm{s} \mathrm{O}_{\mathrm{y}} ?\)

Balance the following equations, and name each reactant and product: (a) \(\mathrm{SF}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{HF}(\ell)\) (b) \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{BF}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HF}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{aq})\)

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