Write chemical equations for the reactions you would expect to occur when (a) \(\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\mathrm{s})\) is heated to a high temperature (b) \(\mathrm{BaCl}_{2}(1)\) is electrolyzed (c) \(\operatorname{Sr}(\text { s) is added to cold dilute } \operatorname{HBr}( \text { aq) }\) (d) \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})\) is added to \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) (e) \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) is heated

Short Answer

Expert verified
The exercise involves writing equations for several types of chemical reactions: (a) thermal decomposition, (b) electrolysis, (c) displacement, (d) neutralization and (e) dehydration.

Step by step solution

01

(a) Heating \(\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\mathrm{s})\)

When \(\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\mathrm{s})\) is heated, it decomposes into magnesium oxide (MgO), water (H2O), and carbon dioxide (CO2). It is a type of thermal decomposition reaction. The equation is:\[\operatorname{Mg}\left(\mathrm{HCO}_{3}\right)_{2}(\mathrm{s})\rightarrow \operatorname{MgO}\left(\mathrm{s}\right)+\mathrm{H}_{2}\mathrm{O}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})\]
02

(b) Electrolysis of \(\mathrm{BaCl}_{2}(l)\)

Electrolysis involves breaking down a compound into its elements using an electric current. \(\mathrm{BaCl}_{2}\) will break down into barium (Ba) and chlorine (Cl2) gas. The equation is:\[\mathrm{BaCl}_{2}(l) \rightarrow \mathrm{Ba}(s)+\mathrm{Cl}_{2}(g)\]
03

(c) Adding $\operatorname{Sr}\left(\mathrm{s}\right)$ to cold dilute $\operatorname{HBr}\left(\mathrm{aq}\right)$

When strontium (Sr) is added to cold dilute hydrobromic acid (HBr), a displacement reaction occurs, creating strontium bromide (SrBr2) and hydrogen gas (H2). The equation is:\[\mathrm{Sr}\left(\mathrm{s}\right)+2 \mathrm{HBr}\left(\mathrm{aq}\right) \rightarrow \mathrm{SrBr}_{2}\left(\mathrm{aq}\right)+\mathrm{H}_{2}\left(\mathrm{g}\right)\]
04

(d) Adding \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})\) to \(\mathrm{H}_{2}\mathrm{SO}_{4}(\mathrm{aq})\)

A neutralization reaction takes place when calcium hydroxide (Ca(OH)2) is added to sulphuric acid (H2SO4), forming calcium sulphate (CaSO4) and water (H2O). The equation is:\[\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{CaSO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\]
05

(e) Heating \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\)

When \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\), or gypsum, is heated, it loses water and forms a plaster of Paris. This is a type of dehydration reaction. The equation is:\[\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{s})+ \frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\]

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!

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

A chemical dictionary gives the following descriptions of the production of some compounds. Write plausible chemical equations based on these descriptions. (a) lead(II) carbonate: adding a solution of sodium bicarbonate to a solution of lead nitrate. (b) lithium carbonate: reaction of lithium oxide with ammonium carbonate solution. (c) hydrogen peroxide: by the action of dilute sulfuric acid on barium peroxide. (d) lead(IV) oxide: action of an alkaline solution of calcium hypochlorite on lead(II) oxide.

The best oxidizing agent of the following oxides is (a) \(\mathrm{Li}_{2} \mathrm{O} ;(\mathrm{b}) \mathrm{MgO} ;(\mathrm{c}) \mathrm{Al}_{2} \mathrm{O}_{3} ;(\mathrm{d}) \mathrm{CO}_{2} ;(\mathrm{e}) \mathrm{SnO}_{2} ;(\mathrm{f}) \mathrm{PbO}_{2}.\)

An analysis of a Solvay-process plant shows that for every \(1.00 \mathrm{kg}\) of \(\mathrm{NaCl}\) consumed, \(1.03 \mathrm{kg}\) of \(\mathrm{NaHCO}_{3}\) are obtained. The quantity of \(\mathrm{NH}_{3}\) consumed in the overall process is \(1.5 \mathrm{kg}.\) (a) What is the percent efficiency of this process for converting NaCl to \(\mathrm{NaHCO}_{3} ?\) (b) Why is so little \(\mathrm{NH}_{3}\) required?

Although the triiodide ion, \(\mathrm{I}_{3}^{-}\), is known to exist in aqueous solutions, the ion is stable in only certain ionic solids. For example, \(\mathrm{CsI}_{3}\) is stable with respect to decomposition to CsI and \(\mathrm{I}_{2},\) but \(\mathrm{LiI}_{3}\) is not stable with respect to LiI and I \(_{2}\). Draw a Lewis structure for the \(I_{3}^{-}\) ion and suggest a reason why \(\mathrm{CsI}_{3}\) is stable with respect to decomposition to the iodide but \(\mathrm{LiI}_{3}\) is not.

Arrange the following compounds in the expected order of increasing solubility in water, and give the basis for your arrangement: \(\mathrm{Li}_{2} \mathrm{CO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}\) \(\mathrm{MgCO}_{3}.\)

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