6 Na (s) + Fe2O3 (s) ⟶ 3 Na2O3 (s) + 2 Fe (s).

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6 Na (s) + Fe2O3 (s) ⟶ 3 Na2O3 (s) + 2 Fe (s). 23. The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflates an automobile airbag: 6 Na (s) + Fe2O3 (s) ⟶ 3 Na2O3 (s) + 2 Fe (s). If 100.0 g of Na and 100.0 g of Fe2O3 are used in this reaction, determine the following: Limiting reactant Reactant in excess Mass of solid iron produced

6 Na (s) + Fe2O3 (s) ⟶ 3 Na2O3 (s) + 2 Fe (s). 100.0 g 100.0 g ? g 6 mol Na 2 mol Fe 1 mol Fe2O3 = 138.0 g Na 111.6 g Fe = 159.6 g Fe2O3 = 100.0 g Na X g Fe = 100.0 g Fe2O3 X = 69.92 g Fe X = 80.87 g Fe Na is in excess Fe2O3 is limiting

24. Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Write a balanced chemical equation for reaction Determine the limiting reactant Determine the excess reactant e) Determine the mass of glucose produced

6 CO2 (g) + 6 H20 (l) → C6H12O6 (aq) + 6 O2 (g) 88.0 g 64.0 g ? g 6 mol CO2 1 mol C6H12O6 6 mol H20 = 264.0 g CO2 180.0 g C6H12O6 = 108.0 g H20 = 88.0 g CO2 X g C6H12O6 = 64.0 g H20 X = 107 g C6H12O6 X = 60.0 g C6H12O6 H20 is in excess CO2 is limiting

28. Aluminum hydroxide (Al(OH)3) is often present in antacids to neutralize stomach acid (HCl). The reaction occurs as follows: Al(OH)3 (s) + 3 HCl (aq) ⟶ AlCl3 (aq) + 3 H2O (l) If 14.0 g of Al(OH)3 is present an antacid tablet determine the theoretical yield of AlCl3 produced when the tablet with HCl. 1 mol Al(OH)3 = 78.0 g Al(OH)3 = 14.0 g Al(OH)3 1 mol AlCl3 133.5 g AlCl3 X g AlCl3 X = 24.0 g AlCl3

Zinc reacts with iodine in a synthesis reaction: 29. Zinc reacts with iodine in a synthesis reaction: Zn + I2 ⟶ ZnI2 a) Determine the theoretical yield if 1.912 mol of zinc is used. 1 mol Zn 1.912 mol ZnI2 319.2 g ZnI2 1 mol ZnI2 1 mol ZnI2 = 610.3 g ZnI2 b) Determine the percent yield if 515.6 g of product is recovered. 515. 6 g ZnI2 = 84.48 % 610.3 g ZnI2

30. Challenge When copper wire is placed into silver nitrate solution, silver crystals and copper(II) nitrate solution form. Write a balanced chemical equation for the reaction If 20.0-g sample of copper is used, determine the theoretical yield of silver. If 60.0 g of silver is recovered from the reaction, determine the percent yield of the reaction.

Cu + 2 AgNO3 ⟶ Cu(NO3)2 + 2 Ag 20.0 g ? g a) Determine the theoretical yield if 20.0 g of Cu is used. 1 mol Cu = 63.5 g Cu = 20.0 g Cu 2 mol Ag 215.8 g Ag X g Ag = 68.0 g Ag b) Determine the percent yield 60.0 g of Ag is recovered. 60.0 g Ag = 88.2 % 68.0g ZnI2