00 mol KOH 2 2 2 OH Cu mol OH Cu g 57 97 KOH mol 2 OH Cu mol 1 293 g CuOH 2 63

00 mol koh 2 2 2 oh cu mol oh cu g 57 97 koh mol 2 oh

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Mass of precipitate = 6.00 mol KOH × 222)OH(Cumol)OH(Cug57.97KOHmol2)OH(Cumol1= 293 g Cu(OH)2 63. M2SO4(aq) + CaCl2(aq) CaSO4(s) + 2 MCl(aq) 1.36 g CaSO4× 44244CaSOmolSOMmol1CaSOg15.136CaSOmol1= 9.99 × 310mol M2SO4From the problem, 1.42 g M2SO4was reacted, so: molar mass = 42342SOMmol1099.9SOMg42.1= 142 g/mol 142 u = 2(atomic mass M) + 32.07 + 4(16.00), atomic mass M = 23 u From periodic table, M = Na (sodium). 64. a. Na+, NO3, Cl, and Ag+ions are present before any reaction occurs. The excess Ag+added will remove all of the Clions present. Therefore, Na+, NO3, and the excess Ag+ions will all be present after precipitation of AgCl is complete. b. Ag+(aq) + Cl(aq) AgCl(s) c. Mass NaCl = 0.641 g AgCl × NaClmolg44.58ClmolNaClmol1AgClmolClmol1g4.143AgClmol1= 0.261 g NaCl Mass % NaCl = mixtureg50.1NaClg261.0× 100 = 17.4% NaCl
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CHAPTER 4 SOLUTION STOICHIOMETRY 111 Acid-Base Reactions65. All the bases in this problem are ionic compounds containing OH-. The acids are either strong or weak electrolytes. The best way to determine if an acid is a strong or weak electrolyte is to memorize all the strong electrolytes (strong acids). Any other acid you encounter that is not a strong acid will be a weak electrolyte (a weak acid), and the formula should be left unaltered in the complete ionic and net ionic equations. The strong acids to recognize are HCl, HBr, HI, HNO3, HClO4, and H2SO4. For the following answers, the order of the equations are formula, complete ionic, and net ionic. a. 2 HClO4(aq) + Mg(OH)2(s) 2 H2O(l) + Mg(ClO4)2(aq) 2 H+(aq) + 2 ClO4(aq) + Mg(OH)2(s) 2 H2O(l) + Mg2+(aq) + 2 ClO4(aq) 2 H+(aq) + Mg(OH)2(s) 2 H2O(l) + Mg2+(aq) b. HCN(aq) + NaOH(aq) H2O(l) + NaCN(aq) HCN(aq) + Na+(aq) + OH(aq) H2O(l) + Na+(aq) + CN(aq) HCN(aq) + OH(aq) H2O(l) + CN(aq) c. HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) H+(aq) + Cl(aq) + Na+(aq) + OH(aq) H2O(l) + Na+(aq) + Cl(aq) H+(aq) + OH(aq) H2O(l) 66. a. 3 HNO3(aq) + Al(OH)3(s) 3 H2O(l) + Al(NO3)3(aq) 3 H+(aq) + 3 NO3(aq) + Al(OH)3(s) 3 H2O(l) + Al3+(aq) + 3 NO3(aq) 3 H+(aq) + Al(OH)3(s) 3 H2O(l) + Al3+(aq) b. HC2H3O2(aq) + KOH(aq) H2O(l) + KC2H3O2(aq) HC2H3O2(aq) + K+(aq) + OH(aq) H2O(l) + K+(aq) + C2H3O2(aq) HC2H3O2(aq) + OH(aq) H2O(l) + C2H3O2(aq) c. Ca(OH)2(aq) + 2 HCl(aq) 2 H2O(l) + CaCl2(aq) Ca2+(aq) + 2 OH(aq) + 2 H+(aq) + 2 Cl(aq) 2 H2O(l) + Ca2+(aq) + 2 Cl(aq) 2 H+(aq) + 2 OH(aq) → 2 H2O(l) or H+(aq) + OH(aq) H2O(l) 67. All the acids in this problem are strong electrolytes (strong acids). The acids to recognize as strong electrolytes are HCl, HBr, HI, HNO3, HClO4, and H2SO4.
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112 CHAPTER 4 SOLUTION STOICHIOMETRY 68. a. Perchloric acid plus potassium hydroxide is a possibility.
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