Chapter 9

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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 110 - <strong>Chapter</strong> 9 - 40. The balanced equation is 16 HCl + 2 KMnO 4 ¡ 5 Cl 2 + 2 KCl + 2 MnCl 2 + 8 H 2 O (a) Reaction between 25 g KMnO 4 and 85 g HCl. Convert each to moles of MnCl 2 . (b) 125 g KMnO 4 2a 1 mol KMnO 4 158.04 g KMnO 4 ba 2 mol MnCl 2 2 mol KMnO 4 b = 0.16 mol MnCl 2 185 g HCl2a 1 mol 36.46 g ba2 mol MnCl 2 16 mol HCl b = 0.29 mol MnCl 2 KMnO 4 is the limiting reactant; 0.16 mol MnCl 2 produced. 175 g KCl2a 1 mol 74.55 g ba8 mol H 2O 2 mol KCl ba18.02 g b = 73 g H mol 2 O (c) Theoretical yield is 91 g Cl 2 ; Percent yield: a 75 g b(100) = 82% yield 91 g (d) Reaction between 25 g HCl and 25 g KMnO 4 . Convert each amount to grams of CL 2 . (e) 1150 g HCl2a 1 mol 36.46 g ba 5 mol Cl 2 16 mol HCl ba70.90 g b = 91 g Cl mol 2 125 g HCl2a 1 mol 36.46 g ba 5 mol Cl 2 16 mol HCl ba70.90 g b = 15 g Cl mol 2 125 g KMnO 4 2a 1 mol 158.04 g ba 5 mol Cl 2 ba 70.90 g b = 28 g Cl 2 mol KMnO 4 mol 2 HCl is the limiting; KMnO 4 is in excess; 15 g Cl 2 will be produced. Calculate the mass of unreacted KMnO 4 : 125 g HCl2a 1 mol . 36.46 g ba2 mol KMnO 4 ba 158.04 g b = 14 g KMnO 16 mol HCl mol 4 will react Unreacted KMnO 4 = 25 g -14 g = 11 g KMnO 4 remain unreacted. 41. The balanced equation is 4Ag + 2 H 2 S + O 2 ¡ 2 Ag 2 S + 2 H 2 O (a) 11.1 g Ag2a 1 mol 107.9 g ba2 mol Ag 2S 4 mol Ag ba247.9 g b = 1.3 g Ag mol 2 S 10.14 g H 2 S2a 1 mol 34.09 g ba2 mol Ag 2S 2 mol H 2 S ba247.9 g b = 1.0 g Ag mol 2 S - 110 -
HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 111 - <strong>Chapter</strong> 9 - (b) H 2 S is limiting 1.0 g Ag 2 S forms. 0.17 g - 0.14 g = 0.03 grams more H 2 S needed to completely react Ag. 42. Mass of the beaker 26.500 g beaker + Ca(OH) 2 26.095 g beaker + CaO 0.405 g H 2 O absorbed 10.405 g H 2 O2a 1 mol 18.02 g b = 2.25 * 10-2 mol H 2 O absorbed Since the reaction is a 1:1 mole, the amount of CaO in the beaker is 2.25 * 10 -2 mol. Convert to grams. 12.25 * 10 -2 56.08 g CaO mol CaO2a b = 1.26 g CaO in the beaker. 26.095 g -1.26 g 24.835 g 10.080 g O 2 2a 1 mol 32.00 g ba2 mol Ag 2S ba 247.9 g b = 1.2 g Ag 1 mol O 2 mol 2 S 11.1 g Ag2a 1 mol 107.9 g ba2 mol H 2S 4 mol Ag ba34.09 g b = 0.17 g H mol 2 S reacts mol beaker + CaO CaO mass of the beaker 43. Pb(NO 3 ) 2 (aq) + 2 KI(aq) ¡ PbI 2 (s) + 2 KNO 3 (aq) (a) The solid is lead (II) iodide, PbI 2 . (b) Double displacement reaction. (c) Calculate the moles of each reactant. [15 g Pb(NO 3 ) 2 ] + a 1 mol 331.2 g b = 0.045 mol Pb(NO 3) 2 115 g KI2a 1 mol b = 0.090 mol KI 166.0 g Stoichiometric quantities of reactants are used. Theoretical yield of PbI 2 is 0.045 mol. Actual yield: Percent yield: a 16.68 g PbI 2 2a 1 mol 461.1 g b = 0.0145 mol PbI 2 0.0145 mol b11002 = 32% yield 0.045 mol - 111 -
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Chapter 9

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