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Chapter 22: Problem 24

Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.

Short Answer

Expert verified
The balanced equations for the given reactions are: (a) \(2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}\) (b) \(H_{2}O + Mg \rightarrow MgO + H_{2}\) (c) \(MnO_{2} + 2H_{2} \rightarrow MnO + 2H_{2}O\) (d) \(CaH_{2} + 2H_{2}O \rightarrow Ca(OH)_{2} + 2H_{2}\)

Step by step solution

01

Writing the unbalanced equation for the reaction involving Aluminum and an acid.

The general equation for the reaction of aluminum (Al) with an acid (HX) is: Al + HX → AlX + H_2 Here, X represents the anion from the acid. Since no specific acid is mentioned, we will choose a common acid for the example, such as hydrochloric acid (HCl).
02

Balancing the equation for Aluminum and hydrochloric acid reaction.

Now we balance the unbalanced equation: Al + HCl → AlCl_3 + H_2 Balancing the elements on both sides: 2Al + 6HCl → 2AlCl_3 + 3H_2 So the balanced equation is: \(2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}\) (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen.
03

Writing the unbalanced equation for the reaction involving steam and magnesium.

The given reaction can be written as: H_2O (steam) + Mg → MgO + H_2
04

Balancing the equation for the steam and magnesium reaction.

Now we balance the unbalanced equation: H_2O + Mg → MgO + H_2 Balancing the elements on both sides: H_2O + Mg → MgO + H_2 So the balanced equation is: \(H_{2}O + Mg \rightarrow MgO + H_{2}\) (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas.
05

Writing the unbalanced equation for the reduction of manganese(IV) oxide.

The given reaction can be written as: MnO_2 + H_2 → MnO + H_2O
06

Balancing the equation for the reduction of manganese(IV) oxide.

Now we balance the unbalanced equation: MnO_2 + H_2 → MnO + H_2O Balancing the elements on both sides: MnO_2 + 2H_2 → MnO + 2H_2O So the balanced equation is: \(MnO_{2} + 2H_{2} \rightarrow MnO + 2H_{2}O\) (d) Calcium hydride reacts with water to generate hydrogen gas.
07

Writing the unbalanced equation for the reaction involving calcium hydride and water.

The given reaction can be written as: CaH_2 + H_2O → Ca(OH)_2 + H_2
08

Balancing the equation for the reaction between calcium hydride and water.

Now we balance the unbalanced equation: CaH_2 + H_2O → Ca(OH)_2 + H_2 Balancing the elements on both sides: CaH_2 + 2H_2O → Ca(OH)_2 + 2H_2 So the balanced equation is: \(CaH_{2} + 2H_{2}O \rightarrow Ca(OH)_{2} + 2H_{2}\)

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Most popular questions from this chapter

Identify the following hydrides as ionic, metallic, or molecular: $(\mathbf{a}) \mathrm{H}_{2} \mathrm{~S},(\mathbf{b}) \mathrm{LiH},(\mathbf{c}) \mathrm{VH}_{0.56}$.

(a) Draw the Lewis structures for at least four species that have the general formula $$[: \mathrm{X} \equiv \mathrm{Y}:]^{n}$$ where \(X\) and Y may be the same or different, and \(n\) may have a value from +1 to \(-2 .\) (b) Which of the compounds is likely to be the strongest Bronsted base? Explain. [Sections 22.1, 22.7, and 22.9]

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Select the member of group 14 that best fits each description: \((\mathbf{a})\) has the greatest tendency to form multiple bonds with itself, (b) forms polymeric structures with oxygen, (c) is a metal with +2 and +4 oxidation states.

Hydrogen peroxide is capable of oxidizing (a) hydrazine to \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O},(\mathbf{b}) \mathrm{SO}_{2}\) to \(\mathrm{SO}_{4}^{2-},(\mathbf{c}) \mathrm{NO}_{2}^{-}\) to \(\mathrm{NO}_{3}^{-},(\mathbf{d}) \mathrm{H}_{2} \mathrm{~S}(g)\) to \(\mathrm{S}(s),(\mathbf{e}) \mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\). Write a balanced net ionic equation for each of these redox reactions.
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