Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Complete and balance the following equations. If no reaction occurs, so state. (a) \(\operatorname{TiCl}_{4}(\mathrm{g})+\mathrm{Na}(1) \stackrel{\Delta}{\longrightarrow}\) (b) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{Ag}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) (d) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq})+\mathrm{KOH}(\mathrm{aq}) \longrightarrow\) (e) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow}\)

Short Answer

Expert verified
The balanced equations are as follows:(a) \( \text{TiCl4(g)} + 4\text{Na(l)} \rightarrow \text{Ti(s)} + 4\text{NaCl(s)} \) (b) \( \text{Cr2O3(s)} + 2\text{Al(s)} \rightarrow \text{Al2O3(s)} + 2\text{Cr(s)} \) (c) No reaction(d) \( \text{K2Cr2O7(aq)} + 2\text{KOH(aq)} \rightarrow \text{2K2CrO4(aq) + H2O(l)} \)(e) \( \text{MnO2(s) + C(s)} \rightarrow \text{Mn(s) + CO2(g)} \)

Step by step solution

01

Complete and balance equation for TiCl4(g) + Na(l)

The reaction is between Titanium(IV) chloride (TiCl4) and Sodium metal (Na). The product of this reaction is Sodium chloride (NaCl) and Titanium (Ti). Therefore the completed chemical equation is \( \text{TiCl4(g)} + 4\text{Na(l)} \rightarrow \text{Ti(s)} + 4\text{NaCl(s)} \)
02

Complete and balance equation for Cr2O3(s) + Al(s)

The reaction between Chromium(III) oxide (Cr2O3) and Aluminum metal (Al) produces Aluminum oxide (Al2O3) and Chromium (Cr). The completed and balanced chemical equation is \( \text{Cr2O3(s)} + 2\text{Al(s)} \rightarrow \text{Al2O3(s)} + 2\text{Cr(s)} \)
03

Complete and balance equation for Ag(s) + HCl(aq)

The reaction is between Silver metal (Ag) and Hydrochloric acid (HCl). However, silver does not react with hydrochloric acid. So, in this case no reaction occurs.
04

Complete and balance equation for K2Cr2O7(aq) + KOH(aq)

This reaction is between Potassium dichromate (K2Cr2O7) and Potassium hydroxide (KOH). The products of this reaction are Potassium chromate (K2CrO4), Water (H2O), and Potassium chloride (KCl). The completed balanced equation is \( \text{K2Cr2O7(aq)} + 2\text{KOH(aq)} \rightarrow \text{2K2CrO4(aq) + H2O(l)} \)
05

Complete and balance equation for MnO2(s) + C(s)

The reaction is between Manganese(IV) oxide (MnO2) and Carbon (C). The products of this reaction will be Manganese (Mn) and Carbon dioxide (CO2). The completed balanced equation is \( \text{MnO2(s) + C(s)} \rightarrow \text{Mn(s) + CO2(g)} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Reactivity Series
Understanding the reactivity series is crucial when predicting the outcomes of chemical reactions. The reactivity series is a list of elements organized by their ability to displace others in a compound. More reactive elements can displace less reactive elements from their compounds during a chemical reaction.

For example, in the exercise, aluminum (Al) is above chromium (Cr) in the reactivity series. This is why aluminum is capable of reducing chromium(III) oxide (Cr2O3) to chromium (Cr) in an oxidation-reduction reaction. The reaction can be represented by the equation \( \text{Cr2O3(s)} + 2\text{Al(s)} \rightarrow \text{Al2O3(s)} + 2\text{Cr(s)} \). Here, aluminum acts as a reducing agent, and chromium(III) oxide is reduced.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, commonly known as redox reactions, are processes that involve the transfer of electrons between elements. When an element gains electrons, it is reduced, and when an element loses electrons, it is oxidized.

In step 1 of the exercise solution, the equation \( \text{TiCl4(g)} + 4\text{Na(l)} \rightarrow \text{Ti(s)} + 4\text{NaCl(s)} \) involves sodium (Na) being oxidized as it loses electrons to form NaCl, whereas titanium tetrachloride (TiCl4) is reduced to titanium (Ti). The ability to correctly balance redox equations is essential in predicting the stoichiometry of the products formed in a reaction.
Chemical Reaction Types
Chemical reactions can be classified into several types, including synthesis, decomposition, single displacement, double displacement, and combustion. This classification helps in understanding the reaction mechanisms and predicting the products.

For instance, the reaction given in step 5, \( \text{MnO2(s) + C(s)} \rightarrow \text{Mn(s) + CO2(g)} \), is a type of redox reaction where manganese dioxide (MnO2) is reduced to manganese (Mn), and carbon (C) is oxidized to carbon dioxide (CO2). This particular reaction is also a decomposition reaction because a single compound (MnO2) breaks down into simpler substances (Mn and CO2). Noting the type of reaction helps in formulating and balancing the chemical equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Both \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) can be used to titrate \(\mathrm{Fe}^{2+}(\mathrm{aq})\) to \(\mathrm{Fe}^{3+}(\mathrm{aq}) .\) Suppose you have available as titrants two solutions: \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}(\mathrm{aq})\). (a) For which solution would the greater volume of titrant be required for the titration of a particular sample of \(\mathrm{Fe}^{2+}(\text { aq }) ?\) Explain. (b) How many \(\mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}(\) aq) would be required for a titration if the same titration requires \(24.50 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) ?\)

The text mentions that scandium metal is obtained from its molten chloride by electrolysis, and that titanium is obtained from its chloride by reduction with magnesium. Why are these metals not obtained by the reduction of their oxides with carbon (coke), as are metals such as zinc and iron?

Write plausible half-equations to represent each of the following in basic solution. (a) oxidation of \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) to \(\mathrm{FeO}_{4}^{2-}\) (b) reduction of \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\) to silver metal

At \(400^{\circ} \mathrm{C}, \Delta G^{\circ}=-25 \mathrm{kJ}\) for the reaction \(2 \mathrm{Hg}(1)+\) \(\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HgO}(\mathrm{s}) .\) If a sample of \(\mathrm{HgO}(\mathrm{s})\) is heated to \(400^{\circ} \mathrm{C},\) what will be the equilibrium partial pressure of \(\mathrm{O}_{2}(\mathrm{g}) ?\)

All of the following elements have an ion displaying the +6 oxidation state except (a) Mo; (b) Cr; (c) Mn; (d) \(\mathrm{V} ;\) (e) S.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free