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Because the oxide ion is basic, metal oxides react readily with acids. (a) Write the net ionic equation for the following reaction: $$ \mathrm{FeO}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (b) Based on the equation in part (a), write the net ionic equation for the reaction that occurs between \(\mathrm{NiO}(s)\) and an aqueous solution of nitric acid.

Short Answer

Expert verified
(a) \( \mathrm{FeO}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \); (b) \( \mathrm{NiO}(s) + 2\mathrm{H}^{+}(aq) \rightarrow \mathrm{Ni}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \).

Step by step solution

01

Identify the Reactants and Products

In the provided reaction \( \mathrm{FeO}(s) + 2 \mathrm{HClO}_{4}(aq) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \), iron(II) oxide (FeO) is the reactant reacting with perchloric acid (HClO₄) to form iron(II) perchlorate \( \mathrm{Fe(ClO}_{4})_{2} \) and water.
02

Write the Complete Ionic Equation

Express all strong electrolytes in ionic form: \( \mathrm{FeO}(s) + 2\mathrm{H}^{+}(aq) + 2\mathrm{ClO}_{4}^{-}(aq) \longrightarrow \mathrm{Fe}^{2+}(aq) + 2\mathrm{ClO}_{4}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \). Iron(II) oxide and water remain in their molecular forms since FeO is a solid and water is a pure liquid.
03

Cancel Spectator Ions

In the complete ionic equation, the \( \mathrm{ClO}_{4}^{-} \) ions appear on both sides of the equation. These ions are spectator ions and can be canceled out. This leaves us with the net ionic equation: \( \mathrm{FeO}(s) + 2\mathrm{H}^{+}(aq) \longrightarrow \mathrm{Fe}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \).
04

Apply the Process to Nickel Oxide and Nitric Acid

Now, apply the steps to the reaction of nickel(II) oxide with nitric acid. \( \mathrm{NiO}(s) + 2\mathrm{HNO}_{3}(aq) \longrightarrow \mathrm{Ni}^{2+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \).
05

Write Complete Ionic Equation for NiO reaction

Write all strong electrolytes in ionic form: \( \mathrm{NiO}(s) + 2\mathrm{H}^{+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) \longrightarrow \mathrm{Ni}^{2+}(aq) + 2\mathrm{NO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \). Both nickel oxide and water remain in molecular form.
06

Cancel Spectator Ions in NiO reaction

Cancel out the spectator ions \( \mathrm{NO}_{3}^{-} \), which appear on both sides of the equation, resulting in the net ionic equation: \( \mathrm{NiO}(s) + 2\mathrm{H}^{+}(aq) \longrightarrow \mathrm{Ni}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \).

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Key Concepts

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

Metal Oxides
Metal oxides are compounds composed of metal elements combined with oxygen. These compounds are typically ionic in nature, meaning they are formed through the transfer of electrons, resulting in positive metal ions and negative oxide ions. Metal oxides often appear as solid substances and can vary in color, depending on the specific metal involved.

One of the key characteristics of metal oxides is their basicity. This means they react with acids to form salt and water, a process that is central to many chemical reactions, particularly in context of acid-base chemistry. For example, when iron(II) oxide, which is a metal oxide, reacts with acids like perchloric acid, it undergoes a chemical reaction to form iron salts and water. These reactions are not only fundamental in understanding how compounds interact with acids but also in explaining corrosion and other processes where metals are involved.
  • Metal oxides: compounds of metals and oxygen
  • Typically solid and ionic
  • React with acids to form salt and water
Understanding the behavior of metal oxides is crucial for comprehending a wide variety of chemical reactions and industrial applications.
Acid-Base Reactions
Acid-base reactions are among the most important types of chemical reactions, playing a crucial role in chemistry. These reactions involve an acid and a base reacting to form a salt and usually water. The basic premise is that the acid donates protons ( hydrogen ions, denoted as 1H^+1), while the base accepts them. This transfer of protons leads to the formation of new products.

When metal oxides react with acids, they do so through an acid-base reaction. In the reaction involving iron(II) oxide (1FeO1) and perchloric acid (1HClO_41), the 1H^+1 ions from the acid react with the oxide ions in the metal oxide, resulting in the formation of water and a corresponding metal salt. This transformation is part of the neutralization process, where the properties of acids and bases are counteracted to yield neutral compounds.
  • Acid-base reactions: involve transfer of protons
  • Result in formation of salt and water
  • Neutralization process
These reactions are foundational in many fields, including analytical chemistry and biochemistry, providing insight into how substances interact at the molecular level.
Spectator Ions
In the realm of chemical reactions, particularly in aqueous solutions, spectator ions can often make understanding the core reactions a bit confusing. Spectator ions are ions that exist in the same form on both the reactant and product sides of a chemical equation. They do not participate directly in the chemical reaction, as they remain unchanged by the end of the reaction process.

In the context of the reaction between iron(II) oxide and perchloric acid, the perchlorate ions (1ClO_4^-1) are spectator ions. In both the complete and net ionic equations, these ions appear in the same form without undergoing any change. They do not affect the outcome of the primary reaction, which involves the formation of water and the corresponding metal cation from the oxide.
  • Spectator ions: unchanged during reaction
  • Present in both reactants and products
  • Do not influence primary reaction outcome
Recognizing and understanding spectator ions is crucial for simplifying equations into net ionic forms, where only the ions participating in the reaction are included. This makes understanding the core chemistry much clearer and helps focus on the main chemical transformations taking place.

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Most popular questions from this chapter

In 2014 , a major chemical leak at a facility in West Virginia released \(28,390 \mathrm{~L}\) of MCHM (4-methylcyclohexylmethanol, \(\mathrm{C}_{8} \mathrm{H}_{16} \mathrm{O}\) ) into the Elk River. The density of MCHM is 0.9074 \(\mathrm{g} / \mathrm{mL}\). (a) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is \(2.00 \mathrm{~m}\) deep, \(90.0 \mathrm{~m}\) wide, and \(90.0 \mathrm{~m}\) long. (b) How much farther down the river would the spill have to spread in order to achieve a "safe" MCHM concentration of \(1.00 \times 10^{-4} \mathrm{M}\) ? Assume the depth and width of the river are constant and the concentration of MCHM is uniform along the length of the spill.

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+& \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\), that must be added to the spill to neutralize the acid?

A \(3.50 \mathrm{~g}\) of an alloy which contains only lead and tin is dissolved in hot \(\mathrm{HNO}_{3} .\) Excess sulfuric acid is added to this solution and \(1.57 g\) of \(\mathrm{PbSO}_{4}(s)\) is obtained. (a) Write the net ionic equation for the formation of \(\mathrm{PbSO}_{4}\). (b) Assuming all the lead in the alloy reacted to form \(\mathrm{PbSO}_{4}\), what was the amount, in grams, of lead and tin in the alloy respectively?

A solid sample of \(\mathrm{Fe}(\mathrm{OH})_{3}\) is added to \(0.500 \mathrm{~L}\) of \(0.250 \mathrm{M}\) aqueous \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The solution that remains is still acidic. It is then titrated with \(0.500 \mathrm{M} \mathrm{NaOH}\) solution, and it takes \(12.5 \mathrm{~mL}\) of the NaOH solution to reach the equivalence point. What mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) was added to the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution?

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for 10.0 min at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) \((\mathbf{c})\) Is this manufacturer in compliance with regulations?

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