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Chapter 5: Problem 70

Following are some laboratory methods occasionally used for the preparation of small quantities of chemicals. Write a balanced equation for each. (a) preparation of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{FeS}(\mathrm{s})\) (b) preparation of \(\mathrm{Cl}_{2}(\mathrm{g}): \mathrm{HCl}(\mathrm{aq})\) is heated with \(\mathrm{MnO}_{2}(\mathrm{s}) ; \mathrm{MnCl}_{2}(\mathrm{aq})\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are other products (c) preparation of \(\mathrm{N}_{2}: \mathrm{Br}_{2}\) and \(\mathrm{NH}_{3}\) react in aqueous solution; \(\mathrm{NH}_{4} \mathrm{Br}\) is another product (d) preparation of chlorous acid: an aqueous suspension of solid barium chlorite is treated with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

Short Answer

Expert verified
The balanced chemical equations are: (a) FeS(s) + 2HCl(aq) → H2S(g) + FeCl2(aq), (b) 4HCl(aq) + MnO2(s) → Cl2(g) + MnCl2(aq) + 2H2O(l), (c) 8NH3 + 3Br2 → N2 + 6NH4Br, and (d) Ba(ClO2)2 + H2SO4 → 2HClO2 + BaSO4(s).

Step by step solution

01

Preparation of H2S(g)

HCl(aq) and FeS(s) react to form H2S(g) and a by-product, FeCl2(s). As per stoichiometry, one molecule of HCl reacts with one molecule of FeS to produce one molecule of H2S and one molecule of FeCl2. Therefore, the balanced chemical equation is: \\[FeS(s) + 2HCl(aq) \longrightarrow H2S(g) + FeCl2(aq)\]
02

Preparation of Cl2(g)

When HCl(aq) is heated with MnO2(s), it leads to the formation of Cl2(g), MnCl2(aq) and H2O(l). Balancing the equation yields: \\[4HCl(aq) + MnO2(s) \longrightarrow Cl2(g) + MnCl2(aq) + 2H2O(l)\]
03

Preparation of N2

NH3 and Br2 react to form N2 and NH4Br. Balancing the equation gives: \\[8NH3 + 3Br2 \longrightarrow N2 + 6NH4Br\]
04

Preparation of chlorous acid

When an aqueous suspension of barium chlorite is treated with dilute H2SO4(aq), chlorous acid and barium sulfate are formed. The balanced equation for this reaction is: \\[Ba(ClO2)2 + H2SO4 \rightarrow 2HClO2 + BaSO4(s)\]

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Most popular questions from this chapter

Explain the important distinctions between (a) a strong electrolyte and strong acid; (b) an oxidizing agent and reducing agent; (c) precipitation reactions and neutralization reactions; (d) half-reaction and overall reaction.

A sample of battery acid is to be analyzed for its sulfuric acid content. A \(1.00 \mathrm{mL}\) sample weighs \(1.239 \mathrm{g}\). This \(1.00 \mathrm{mL}\) sample is diluted to \(250.0 \mathrm{mL}\), and \(10.00 \mathrm{mL}\) of this diluted acid requires \(32.44 \mathrm{mL}\) of \(0.00498 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. What is the mass percent of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the battery acid? (Assume that complete ionization and neutralization of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) occurs.)

A piece of marble (assume it is pure \(\mathrm{CaCO}_{3}\) ) reacts with \(2.00 \mathrm{L}\) of \(2.52 \mathrm{M} \mathrm{HCl}\). After dissolution of the marble, a \(10.00 \mathrm{mL}\) sample of the resulting solution is withdrawn, added to some water, and titrated with 24.87 mL of 0.9987 M NaOH. What must have been the mass of the piece of marble? Comment on the precision of this method; that is, how many significant figures are justified in the result?

Sodium hydroxide used to make standard \(\mathrm{NaOH}(\mathrm{aq})\) solutions for acid-base titrations is invariably contaminated with some sodium carbonate. (a) Explain why, except in the most precise work, the presence of this sodium carbonate generally does not seriously affect the results obtained, for example, when \(\mathrm{NaOH}(\mathrm{aq})\) is used to titrate HCl(aq). (b) Conversely, show that if \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) comprises more than \(1 \%\) to \(2 \%\) of the solute in NaOH(aq), the titration results are affected.

Household ammonia, used as a window cleaner and for other cleaning purposes, is \(\mathrm{NH}_{3}(\mathrm{aq}) .\) The \(\mathrm{NH}_{3}\) present in a 5.00 \(\mathrm{mL}\) sample is neutralized by 28.72 \(\mathrm{mL}\) of \(1.021 \mathrm{M} \mathrm{HCl}\). The net ionic equation for the neutralization is \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})\) What is the molarity of \(\mathrm{NH}_{3}\) in the sample?
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