Chapter 9

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$Math 095 Final Exam Review - Faculty.chemeketa.edu$

HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 114 - <strong>Chapter</strong> 9 - (b) 475 g C 2 H 5 OH represents 84.6% of the theoretical yield. Calculate the theoretical yield. theoretical yield = 475 g 0.846 = 561 g C 2H 5 OH Now calculate the g C 6 H 12 O 6 needed to produce 561 g C 2 H 5 OH. 1561 g C 2 H 5 OH2a 1 mol 46.07 g b ¢ 1 mol C 6H 12 O 6 2 mol C 2 H 5 OH ≤ a 180.2 g b = 1.10 * 10 3 g C mol 6 H 12 O 6 49. The balanced equations are: CaCl 2 (aq) + 2 AgNO 3 (aq) ¡ Ca(NO 3 ) 2 (aq) + 2 AgCl(s) MgCl 2 (aq) + 2 AgNO 3 (aq) ¡ Mg1NO 3 2 2 (aq) + 2 AgCl(s) 1 mol of each salt will produce the same amount (2 mol) of AgCl. MgCl 2 has a higher percentage of Cl than CaCl 2 because Mg has a lower atomic mass than Ca. Therefore, on an equal mass basis, MgCl 2 will produce more AgCl than will CaCl 2 . Calculations show that 1.00 g MgCl 2 produces 3.01 g AgCl, and 1.00 g CaCl 2 produces 2.56 g AgCl. 50. The balanced equation is Li 2 O + H 2 O ¡ 2 LiOH The conversion is: g H 2 O ¡ mol H 2 O ¡ mol Li 2 O ¡ g Li 2 O ¡ kg Li 2 O ¢ 2500 g H 2O astronaut day ≤ a 1 mol 18.02 g b ¢ 1 mol Li 2O ¢ 4.1 kg Li 2O astronaut day ≤130 days213 astronauts2 = 3.7 * 102 kg Li 2 O 51. The balanced equation is H 2 SO 4 + 2 NaCl ¡ Na 2 SO 4 + 2 HCl First calculate the g HCl to be produced 120.0 L HCl solution2a Then calculate the g H 2 SO 4 required to produce the HCl 11.01 * 10 4 g HCl2a 1 mol 36.46 g b ¢ 1 mol H 2SO 4 ≤ a 98.09 g 2 mol HCl 1 mol b = 1.36 * 104 g H 2 SO 4 Finally, calculate the kg H 2 SO 4 (96%) 1 mol H 2 O ≤ a 29.88 g mol ba 1 kg 1000 g b = 4.1 kg Li 2O astronaut day 1000 mL ba 1.20 g 1 L 1.00 mL b10.4202 = 1.01 * 104 g HCl 11.36 * 10 4 g H 2 SO 4 2¢ 1.00 g H 2SO 4 solution 0.96 g H 2 SO 4 ≤ a 1 kg 1000 g b = 14 kg concentrated H 2SO 4 - 114 -
HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 115 - <strong>Chapter</strong> 9 - 52. Percent yield of H 2 SO 4 1100.0 g S2a 1 mol b = 3.118 mol S to start with 32.07 g 3.118 mol S ¡ 3.118 mol SO 2 - 10% = 2.806 mol SO 2 -0.3118 2.8062 2.806 mol SO 2 ¡ 2.806 mol SO 3 - 10% = 2.525 mol SO 3 -0.2806 2.5254 2.525 mol SO 3 ¡ 2.525 mol H 2 SO 4 - 10% = 2.273 mol H 2 SO 4 -0.2525 2.2725 12.273 mol H 2 SO 4 2a 98.09 g b = 223.0 g H mol 2 SO 4 formed 13.118 mol S2a 1 mol H 2SO 4 b = 3.118 mol H 1 mol S 2 SO 4 (theoretical yield) ¢ 2.273 mol H 2SO 4 3.118 mol H 2 SO 4 ≤11002 = 72.90% yield Alternate Solution: Calculation of yield. There are three chemical steps to the formation of H 2 SO 4 . Each step has a 10% loss of yield. Step 1: Step 2: Step 3: 100% yield - 10% = 90.00% yield 90.00% yield - 10% = 81.00% yield 81.00% yield - 10% = 72.90% yield Now calculate the grams of product. One mole of sulfur will yield a maximum of 1 mol H 2 SO 4 . Therefore 3.118 mol S will give a maximum of 3.118 mol H 2 SO 4 . 13.118 mol S2a 1 mol H 2SO 4 1 mol S ba 98.09 g b10.72902 = 223.0 g H mol 2 SO 4 yield 53. According to the equations, the moles of CO 2 come from both reactions and the moles H 2 O come from only the first reaction. So the mol NaHCO 3 = 2 * mol H 2 O = 2 * 0.0357 mol = 0.0714 mol NaHCO 3 10.0714 mol NaHCO 3 2a 84.01 g b = 6.00 g NaHCO mol 3 in the sample 110.00 g NaHCO 3 + Na 2 CO 3 2 - 6.00 g NaHCO 3 = 4.00 g Na 2 CO 3 in the sample - 115 -
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Chapter 9

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