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Write chemical equations for the following reactions: (a) the displacement of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{HCl}(\mathrm{aq})\) by \(\mathrm{Al}(\mathrm{s})\) (b) the re-forming of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) with steam (c) the reduction of \(\mathrm{MnO}_{2}(\mathrm{s})\) to \(\mathrm{Mn}(\mathrm{s})\) with \(\mathrm{H}_{2}(\mathrm{g})\)

Short Answer

Expert verified
The balanced chemical equations are: (a) \(2Al(s) + 6HCl(aq) \rightarrow 2AlCl_{3}(aq) + 3H_{2}(g)\), (b) \(C_{3}H_{8}(g) + 4H_{2}O(g) \rightarrow 3CO_{2}(g) + 8H_{2}(g)\), and (c) \(MnO_{2}(s) + H_{2}(g) \rightarrow Mn(s) + 2H_{2}O(l)\)

Step by step solution

01

Displacement of \(H_{2}(g)\) from \(HCl(aq)\) by \(Al(s)\)

In this displacement reaction, Aluminum (Al) replaces Hydrogen (H) in Hydrochloric acid (HCl) to form Aluminum Chloride (AlCl3) and Hydrogen (H2) gas. The balanced equation is:\(2Al(s) + 6HCl(aq) \rightarrow 2AlCl_{3}(aq) + 3H_{2}(g)\)
02

Reforming of Propane gas \((C_{3}H_{8})\) with Steam

When propane gas reacts with steam, it forms Carbon Dioxide (CO2) and Hydrogen gas (H2). The balanced equation is:\(C_{3}H_{8}(g) + 4H_{2}O(g) \rightarrow 3CO_{2}(g) + 8H_{2}(g)\)
03

Reduction of \(MnO_{2}(s)\) to \(Mn(s)\) with \(H_{2}(g)\)

In this reduction reaction, Hydrogen (H2) gas reduces Manganese Dioxide (MnO2) to Manganese (Mn). The balanced equation is: \(MnO_{2}(s) + H_{2}(g) \rightarrow Mn(s) + 2H_{2}O(l)\)

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Key Concepts

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

Displacement Reactions
Displacement reactions, also known as single-replacement reactions, are a fascinating type of chemical reaction where one element replaces another in a compound. Imagine a scenario where a new dance partner arrives and takes over for your current partner. This is similar to how a displacement reaction works in chemistry.
The general form for these reactions can be shown as:
  • A + BC → AC + B
In the given problem, Aluminum (Al), which is quite reactive, displaces Hydrogen (H) from Hydrochloric acid (HCl). This transformation results in Aluminum Chloride (AlCl₃) and Hydrogen gas (H₂), both of which are different chemical species than what we started with.
Here's the precise balanced equation for this process:
  • 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
Ensuring the equation is balanced is essential because it follows the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system. This means all the atoms in the reactants must be accounted for in the products.
Reforming Reactions
Reforming reactions are chemical processes in which chemical compounds are rearranged to form new compounds. These are crucial in industries that depend on hydrocarbon fuels, such as petrochemicals. Picture a set of blocks that can be rearranged to create completely different structures, similar to what happens in reforming reactions.
An excellent example of this reaction type is the reforming of propane gas defined by the equation:
  • C₃H₈(g) + 4H₂O(g) → 3CO₂(g) + 8H₂(g)
During this reaction, propane reacts with steam under specific conditions, producing carbon dioxide (CO₂) and hydrogen gas (H₂) among other potential products. These reactions typically require a catalyst, which speeds up the reaction without being consumed in the process.
Understanding reforming reactions allows us to convert raw materials into usable fuel for a variety of applications, showing their importance in energy and industrial chemistry.
Reduction Reactions
Reduction reactions are a key concept in chemistry where a material gains electrons. In essence, it is the opposite of oxidation. Together, these reactions are part of the redox (reduction-oxidation) processes.
Whenever you hear a buzz about electrons being transferred, a redox reaction is likely happening. In the reduction phase, the element being reduced, gains electrons, thus reducing its oxidation state.
In the featured exercise, Manganese Dioxide (MnO₂) is subjected to a reduction process with hydrogen gas (H₂). Here's how it unfolds chemically:
  • MnO₂(s) + H₂(g) → Mn(s) + 2H₂O(l)
Here, hydrogen acts as the reducing agent by donating electrons, which results in the reduction of MnO₂ to Mn, while hydrogen itself gets oxidized to water (H₂O).
This reduction reaction is essential in processes such as metal extraction, where pure metals are obtained from their ores. Such reactions are widely utilized across various fields, from metallurgy to synthetic organic chemistry.

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Most popular questions from this chapter

Joseph Priestley, a British chemist, was credited with the discovering oxygen in \(1774 .\) In his experiments, he generated oxygen gas by heating \(\mathrm{HgO}(\mathrm{s}) .\) The other product of the decomposition reaction is \(\mathrm{Hg}(1) .\) What volume of wet \(\mathrm{O}_{2}(\mathrm{g})\) is obtained from the decomposition of \(1.0 \mathrm{g} \mathrm{HgO}(\mathrm{s}),\) if the gas is collected over water at \(25^{\circ} \mathrm{C}\) and a barometric pressure of \(756 \mathrm{mmHg} ?\) The vapor pressure of water is \(23.76 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C}\).

The solubility of \(\mathrm{Cl}_{2}(\mathrm{g})\) in water is \(6.4 \mathrm{g} \mathrm{L}^{-1}\) at \(25^{\circ} \mathrm{C}\) Some of this chlorine is present as \(\mathrm{Cl}_{2},\) and some is found as HOCl or Cl^- For the hydrolysis reaction $$\begin{array}{c}\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \\ \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\\K_{c}=4.4 \times 10^{-4}\end{array}$$ For a saturated solution of \(\mathrm{Cl}_{2}\) in water, calculate \(\left[\mathrm{Cl}_{2}\right],[\mathrm{HOCl}],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) and \(\left[\mathrm{Cl}^{-}\right]\).

What mass of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) was present in a sample that required \(26.50 \mathrm{mL}\) of \(0.0510 \mathrm{M} \mathrm{KMnO}_{4}\) for its oxidation to \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in an acidic solution? \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\).

All the group 15 elements form trifluorides, but nitrogen is the only group 15 element that does not form a pentafluoride. (a) Suggest a reason why nitrogen does not form a pentafluoride. (b) The observed bond angle in \(\mathrm{NF}_{3}\) is approximately \(102.5^{\circ} \mathrm{C} .\) Use VSEPR theory to rationalize the structure of the \(\mathrm{NF}_{3}\) molecule.

Use the following electrode potential diagram for basic solutions to classify each of the statements below as true or false. Assume standard conditions. $$\begin{aligned}\mathrm{SO}_{4}^{2-} \stackrel{-0.936 \mathrm{V}}{\longrightarrow} \mathrm{SO}_{3}^{2-} & \stackrel{-0.576 \mathrm{V}}{\longrightarrow} \\\& \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \stackrel{-0.74 \mathrm{V}}{\longrightarrow} \mathrm{S} \stackrel{-0.476 \mathrm{V}}{\longrightarrow} \mathrm{S}^{2-}\end{aligned}$$ (a) Sulfate \(\left(\mathrm{SO}_{4}^{2-}\right)\) is a stronger oxidant than thiosulfate \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2}\right)\) in basic solution. (b) \(S^{2-}\) can be used as a reducing agent in basic solutions. (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is stable with respect to disproportionation to \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\) in basic solution.

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