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Mercury(II) nitrate is treated with hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) forming a precipitate and a solution. Write the molecular equation and the net ionic equation for the reaction. An acid is formed; is it strong or weak? Name each of the products. If \(65.65 \mathrm{~g}\) of mercury(II) nitrate and \(4.26 \mathrm{~g}\) of hydrogen sulfide are mixed in \(395.0 \mathrm{~g}\) of water to form \(54.16 \mathrm{~g}\) of precipitate, what is the mass of the solution after the reaction?

Short Answer

Expert verified
Molecular equation: \( \text{Hg(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{HgS (s)} + 2\text{HNO}_3 \); Net ionic equation: \( \text{Hg}^{2+} + \text{S}^{2-} \rightarrow \text{HgS (s)} \). Acid formed is strong. Mass of solution: 410.75 g.

Step by step solution

01

Write the Molecular Equation

The reaction between mercury(II) nitrate, \( \text{Hg(NO}_3\text{)}_2 \), and hydrogen sulfide, \( \text{H}_{}2\text{S} \), can be expressed as: \[ \text{Hg(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{HgS (s)} + 2\text{HNO}_3 \]. Here, mercury(II) sulfide (\( \text{HgS} \)) is the precipitate, and nitric acid (\( \text{HNO}_3 \)) is the solution product.
02

Determine Acid Strength

Nitric acid (\( \text{HNO}_3 \)) is the acid formed in this reaction. \( \text{HNO}_3 \) is known to be a strong acid because it completely dissociates in water to form hydrogen ions.
03

Write the Net Ionic Equation

To write the net ionic equation, focus on the ions involved in the formation of the solid: \[ \text{Hg}^{2+} + \text{S}^{2-} \rightarrow \text{HgS (s)} \]. This represents the chemical change and indicates the formation of the insoluble precipitate, mercury(II) sulfide.
04

Calculate Mass of Solution After Reaction

The initial mass consists of the mercury(II) nitrate, hydrogen sulfide, and water: \[ 65.65 \text{ g} + 4.26 \text{ g} + 395.0 \text{ g} = 464.91 \text{ g} \]. Subtract the mass of the precipitate: \[ 464.91 \text{ g} - 54.16 \text{ g} = 410.75 \text{ g} \]. Therefore, the mass of the solution after the reaction is \( 410.75 \text{ g} \).

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Key Concepts

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

Molecular Equation
A molecular equation is a representation of a chemical reaction where each compound is listed with its neutral formula, indicating the substances involved in the reaction process. In this specific exercise, where mercury(II) nitrate reacts with hydrogen sulfide, the molecular equation is represented as:\[ \text{Hg(NO}_3\text{)}_2 + \text{H}_2\text{S} \rightarrow \text{HgS (s)} + 2\text{HNO}_3 \]Here is what each part means:
  • Hg(NO3)2: Mercury(II) nitrate, a salt that disassociates in water into ions.
  • H2S: Hydrogen sulfide, a sulfide source here.
  • HgS: Mercury(II) sulfide, the solid product or precipitate formed in the reaction.
  • HNO3: Nitric acid, formed in the aqueous solution.
This equation provides an overall look at the substances before and after the reaction, covering the state (solid or aqueous) of the products.
Net Ionic Equation
The net ionic equation portrays the essence of a chemical reaction by focusing solely on the components that undergo a change. To derive this equation, we focus on the dissolved ions that form the insoluble product. Here, mercury ions and sulfide ions combine to form mercury(II) sulfide, represented as:\[ \text{Hg}^{2+} + \text{S}^{2-} \rightarrow \text{HgS (s)} \]In this equation:
  • Hg2+: Mercury(II) cations from the dissociation of mercury(II) nitrate in solution.
  • S2-: Sulfide anions that originate from hydrogen sulfide.
  • HgS (s): Mercury(II) sulfide, the insoluble product that precipitates out of the solution.
This equation effectively captures the essence of the precipitation reaction, omitting the spectator ions that are not directly involved in forming the precipitate.
Acid Strength
In this reaction, nitric acid ( HNO3 ) is produced as one of the products dissolved in the solution. Understanding acid strength is essential to predict its behavior in aqueous solutions. Nitric acid is classified as a strong acid. Characteristics of a strong acid like HNO3:
  • It dissociates completely in water, releasing hydrogen ions ( H+ ).
  • Strong acids have a high capacity to donate protons, making them efficient in reactions that require a proton donor.
Knowing that nitric acid is a strong acid can help predict its impact and role in subsequent reactions where it is present. Complete dissociation means more ions in the solution, influencing the reaction dynamics and properties of the solution.
Mass Calculation
Understanding how to calculate the mass of the solution after a reaction helps assess the physical changes that occur, particularly when a precipitate forms. In this exercise, initially, you have the masses of each reactant plus water, which are:
  • Mercury(II) nitrate: 65.65 g
  • Hydrogen sulfide: 4.26 g
  • Water: 395.0 g
Adding these values gives the total initial mass:\[ 65.65 \text{ g} + 4.26 \text{ g} + 395.0 \text{ g} = 464.91 \text{ g} \]After the reaction, mercury(II) sulfide precipitates, and you need to subtract its mass from the total:\[ 464.91 \text{ g} - 54.16 \text{ g} = 410.75 \text{ g} \]Thus, the mass of the remaining solution, containing dissolved substances like nitric acid, is 410.75 g. Knowing the mass of the solution helps in determining the solution's composition and potentially its concentration.

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Most popular questions from this chapter

Elemental bromine is the source of bromine compounds. The element is produced from certain brine solutions that occur naturally. These brines are essentially solutions of calcium bromide that, when treated with chlorine gas, yield bromine in a displacement reaction. What are the molecular equation and net ionic equation for the reaction? A solution containing \(40.0 \mathrm{~g}\) of calcium bromide requires \(14.2 \mathrm{~g}\) of chlorine to react completely with it, and \(22.2 \mathrm{~g}\) of calcium chloride is produced in addition to whatever bromine is obtained. How many grams of calcium bromide are required to produce 10.0 pounds of bromine?

You have a sample of a rat poison whose active ingredient is thallium(I) sulfate. You analyze this sample for the mass percentage of active ingredient by adding potassium iodide to precipitate yellow thallium(I) iodide. If the sample of rat poison weighed \(759.0 \mathrm{mg}\) and you obtained \(212.2 \mathrm{mg}\) of the dry precipitate, what is the mass percentage of the thallium(I) sulfate in the rat poison?

What volume of \(0.230 \mathrm{M} \mathrm{HNO}_{3}\) (nitric acid) reacts with \(49.0 \mathrm{~mL}\) of \(0.330 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (sodium carbonate) in the following reaction? \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) \(2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\)

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Decide whether a reaction occurs for each of the following. If it does not, write \(N R\) after the arrow. If it does, write the balanced molecular equation; then write the net ionic eauation. (a) \(\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{HNO}_{3} \longrightarrow\) (b) \(\mathrm{NaBr}+\mathrm{HClO}_{4} \longrightarrow\) (c) \(\mathrm{CaCl}_{2}+\mathrm{NaNO}_{3} \longrightarrow\) (d) \(\mathrm{MgSO}_{4}+\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \longrightarrow\)

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