5 2 2 7 2 5 2 2 7 2 00429 g C 2 H 5 OH blood g 30 OH H C g 0429 5 2 u 100 0143

5 2 2 7 2 5 2 2 7 2 00429 g c 2 h 5 oh blood g 30 oh

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5227252272= 0.0429 g C2H5OH bloodg0.30OHHCg0429.052u100 = 0.143% C2H5OH 97. a. Let x= mass of Mg, so 10.00 ±x= mass of Zn. Ag+(aq) + Cl±(aq) oAgCl(s). From the given balanced equations, there is a 2 : 1 mole ratio between mol Mg and mol Cl±. The same is true for Zn. Because mol Ag+= mol Cl±present, one can setup an equation relating mol Cl±present to mol Ag+added. b. 0.156 L × ²xg Mg × ZnmolClmol2Zng38.65Znmol1Zng)00.10(MgmolClmol2Mgg31.24Mgmol1±±uu±²ux= 0.156 L × ²±²uAgmolClmol1LAgmol00.3= 0.468 mol Cl±38.65)00.10(231.242xx±²= 0.468, 24.31 × 65.38¸¹·¨©§±²468.038.65200.2031.242xx(130.8)x+ 486.2 ±(48.62)x= 743.8 (carrying 1 extra significant figure) (82.2)x= 257.6, x= 3.13 g Mg; % Mg = mixtureg00.10Mgg13.3× 100 = 31.3% Mg ±²uAgmolClmol1LAgmol00.3= 0.468 mol Cl±= 0.468 mol HCl added MHCl= L0.0780mol0.468= 6.00 MHCl
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92 CHAPTER 4 SOLUTION STOICHIOMETRY 98. Let x= mass of NaCl, and let y= mass K2SO4. So x+ y= 10.00. Two reactions occur: Pb2+(aq) + 2 Cl±(aq) oPbCl2(s) and Pb2+(aq) + SO42±(aq) oPbSO4(s) Molar mass of NaCl = 58.44 g/mol; molar mass of K2SO4= 174.27 g/mol; molar mass of PbCl2= 278.1 g/mol; molar mass of PbSO4= 303.3 g/mol 44.58x= moles NaCl; 27.174y= moles K2SO4mass of PbCl2+ mass PbSO4= total mass of solid 44.58x(1/2)(278.1) + 27.174y(303.3) = 21.75 We have two equations: (2.379)x+ (1.740)y= 21.75 and x+ y= 10.00. Solving: x= 6.81 g NaCl; mixtureg00.10NaClg81.6u100 = 68.1% NaCl 99. Zn(s) + 2 AgNO2(aq) o2 Ag(s) + Zn(NO2)2(aq) Let x= mass of Ag and y= mass of Zn after the reaction has stopped. Then x+ y= 29.0 g. Because the moles of Ag produced will equal two times the moles of Zn reacted: (19.0 ±y) g Zn Agg9.107Agmol1AggZnmol1Agmol2Zng38.65Znmol1uuuxSimplifying: 3.059 × 10±2(19.0 ±y) = (9.268 × 10±3)xSubstituting x= 29.0 ±yinto the equation gives: 3.059 × 10±2(19.0 ±y) = 9.268 × 10±3(29.0 ±y) Solving: 0.581 ±(3.059 × 10±2)y= 0.269 ±(9.268 × 10±3)y, (2.132 × 10±2)y= 0.312, y= 14.6 g Zn 14.6 g Zn are present, and 29.0 ±14.6 = 14.4 g Ag are also present after the reaction is stopped. 100. 0.2750 L × 0.300 mol/L = 0.0825 mol H+; let y= volume (L) delivered by Y and z= volume (L) delivered by Z.
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CHAPTER 4 SOLUTION STOICHIOMETRY 93 H+(aq) + OH±(aq) oH2O(l); = 0.0825 mol H+ 0.2750 L + y+ z= 0.655 L, y+ z= 0.380, z= 0.380 ±yy(0.150) + (0.380 ±y)(0.250) = 0.0825, solving: y= 0.125 L, z= 0.255 L Flow rate for Y = min65.60mL125= 2.06 mL/min; flow rate for Z = min65.60mL255= 4.20 mL/min 101. Molar masses: KCl, 39.10 + 35.45 = 74.55 g/mol; KBr, 39.10 + 79.90 = 119.00 g/mol, AgCl, 107.9 + 35.45 = 143.4 g/mol; AgBr, 107.9 + 79.90 = 187.8 g/mol Let x= number of moles of KCl in mixture and y = number of moles of KBr in mixture. Ag++ Cl±oAgCl and Ag++ Br±oAgBr; so, x= moles AgCl and y= moles AgBr. Setting up two equations from the given information: 0.1024 g = (74.55)x+ (119.0)yand 0.1889 g = (143.4)x+ (187.8)yMultiply the first equation by 0.1198.187, and then subtract from the second. 0.1889 = (143.4)x+ (187.8)y±0.1616 = ±(117.7)x±(187.8)y0.0273 = (25.7)x, x= 1.06 × 10±3mol KCl 1.06 × 10±3mol KCl × KClmolKClg74.55= 0.0790 g KCl Mass % KCl = g1024.0g0790.0× 100 = 77.1%, % KBr = 100.0 ±77.1 = 22.9% 102. Pb2+(aq) + 2 Cl±(aq) oPbCl2(s) 3.407 g PbCl2× 2222PbClmolPbmol1PbClg1.278PbClmol1²u= 0.01225 mol Pb2+L1000.2mol01225.0
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