Chapter5-HW-S07

Chapter5-HW-S07 - Chapter 5 Homework Solutions 15(a Possible exchange products are MnS(Table 5.1 page 166 Rule 13 Insoluble solid precipitate

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Chapter 5 Homework Solutions 15 . (a) Possible exchange products are: MnS ( Table 5.1 , page 166, Rule 13) Insoluble solid precipitate NaCl (Rule 1) Soluble MnCl 2 (aq) + Na 2 S(aq) MnS(s) + 2 NaCl(aq) (b) Possible exchange products are: Cu(NO 3 ) 2 (Rule 2) Soluble H 2 SO 4 (Table 5.2) Soluble strong electrolyte HNO 3 (aq) + CuSO 4 (aq) “N.R.” (c) Possible exchange products are: H 2 O Molecular liquid; no significant ionization NaClO 4 (Rule 1) Soluble NaOH(aq) + HClO 4 (aq) “N.R.” (d) Possible exchange products are: HgS (Rule 13) Insoluble solid precipitate NaNO 3 (Rule 1) Soluble Hg(NO 3 ) 2 (aq) + Na 2 S(aq) HgS(s) + 2 NaNO 3 (aq) (e) Possible exchange products are: PbCl 2 (Rule 3 exception) Insoluble solid precipitate HNO 3 (Table 5.2) Soluble strong electrolyte Pb(NO 3 ) 2 (aq) + 2 HCl(aq) PbCl 2 (s) + 2 HNO 3 (aq) (f) Possible exchange products are: BaSO 4 (Rule 4 exception) Insoluble solid precipitate HCl (Table 5.2) Soluble strong electrolyte BaCl 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2 HCl(aq) 16 . The 2 NO 3 ions are found both in the reactants and the products of the complete ionic equation; therefore, NO 3 , nitrate ion, is a spectator ion. Everything else changes in one way or another. This is an example of a neutralization reaction that assists in the dissolving of an insoluble solid. 2 H + (aq) + Mg(OH) 2 (s) 2 H 2 O( l ) + Mg 2+ (aq) This is an exchange reaction. 20. All ions have aqueous phase in the equations below. Potassium sulfide is K 2 S and iron(III) chloride is FeCl 3 . The cross combinations are iron(III) sulfide, Fe 2 S 3 , which is insoluble (according to rule 12: All sulfides are insoluble…) and potassium chloride, KCl, which is soluble (according to rule 1: All Group 1A… ion compounds are soluble.) balanced overall equation: 3 K 2 S (aq) + 2 FeCl 3 (aq) Fe 2 S 3 (s) + 6 KCl(aq) 6 K + + 3 S 2– + 2 Fe 3+ + 6 Cl Fe 2 S 3 (s) + 6 K + + 6 Cl complete ionic equation Eliminate spectator ions, K + , potassium ions and Cl , chlroide ions.
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3 S 2– + 2 Fe 3+ Fe 2 S 3 (s) net ionic equation The precipitate is iron (III) sulfide. 25 . Barium hydroxide is Ba(OH) 2 , a soluble hydroxide compound (Table 5.1, Rule 10). Nitric acid is HNO 3 , a strong acid (Table 5.2). Each OH ion in the solution reacts with one H + ion to make one H 2 O molecule. The ionic compound produced during this neutralization is Ba(NO 3 ) 2 . It is soluble (Table 5.1, Rule 10). Ba(OH) 2 (aq) + 2 HNO 3 (aq) Ba(NO 3 ) 2 (aq) + 2 H 2 O( l ) balanced 32 . (a) KOH is a base, producing K + and OH ions when dissolved in water. (b) Mg(OH) 2 is a base, producing Mg 2+ and OH ions when dissolved in water. (c) HClO is an acid, producing H + and ClO ions when dissolved in water. (d) HBr is an acid, producing H + and Br ions when dissolved in water. (e) LiOH is a base, producing Li + and OH ions when dissolved in water. (f) H 2 SO 3 is an acid, producing H + , HSO 3 , and SO 3 2– ions when dissolved in water.
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This note was uploaded on 09/13/2008 for the course CHEM 1410 taught by Professor Acree during the Spring '08 term at North Texas.

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Chapter5-HW-S07 - Chapter 5 Homework Solutions 15(a Possible exchange products are MnS(Table 5.1 page 166 Rule 13 Insoluble solid precipitate

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