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HW Answers

HW Answers. 10.18 & 19. Assignment. Read pp. 306-307, practice problems #1-4 on p. 307; problems p. 334 #1 & 3. p. 307 #1. Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 How many grams of Al are needed to completely react with 135 g Fe 2 O 3 ? Conversion: 135 g Fe 2 O 3 to grams of Al.

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HW Answers

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  1. HW Answers 10.18 & 19

  2. Assignment • Read pp. 306-307, practice problems #1-4 on p. 307; problems p. 334 #1 & 3

  3. p. 307 #1 Fe2O3 + 2Al → 2Fe + Al2O3 • How many grams of Al are needed to completely react with 135 g Fe2O3? • Conversion: 135 g Fe2O3 to grams of Al This is the answer. This is the amount given in the question. 135gFe2O3 45.6gAl Divide by molar mass of Fe2O3 (56x2 + 16x3 = 160 g/mol) Multiply by molar mass of Al (27 g/mol) 2molAl___ 1molFe2O3 = 0.84 molFe2O3 X 1.69molAl This is the mole ratio from the balanced equation.

  4. p. 307 #2 Fe2O3 + 2Al → 2Fe + Al2O3 • How many grams of Al2O3can form when 23.6 g Al react with excess Fe2O3? • Conversion: 23.6 g Al to grams of Al2O3 This is the answer. This is the amount given in the question. 23.6 g Al 44.9 gAl2O3 Divide by molar mass of Al(27 g/mol) Multiply by molar mass of Al2O3 (27x2 + 16x3 = 102 g/mol) 1molAl2O3 2molAl = 0.87 molAl X 0.44molAl2O3 This is the mole ratio from the balanced equation.

  5. p. 307 #3 Fe2O3 + 2Al → 2Fe + Al2O3 • How many grams of Fe2O3react with excess Al to make 475 g Fe? Conversion: 475 g Fe to grams of Fe2O3 This is the answer. This is the amount given in the question. 475 g Fe 679 gFe2O3 Divide by molar mass of Fe(56 g/mol) Multiply by molar mass of Fe2O3 (56x2 + 16x3 = 160 g/mol) 1molFe2O3 2 molFe = 8.48 molFe X 4.24 molFe2O3 This is the mole ratio from the balanced equation.

  6. p. 307 #4 Fe2O3 + 2Al → 2Fe + Al2O3 • How many grams of Fe will form when 97.6 g Al2O3 form? • Conversion: 97.6 g Al2O3to grams of Fe This is the answer. This is the amount given in the question. 97.6 g Al2O3 107 gFe Divide by molar mass of Al2O3 (27x2 + 16x3 = 102 g/mol) Multiply by molar mass of Fe (56 g/mol) 2molFe___ 1molAl2O3 = 0.957 molAl2O3 X 1.91molFe This is the mole ratio from the balanced equation.

  7. 334 #1 • Using the mole ratio of the reactants and products in a chemical reaction, what will you most likely be able to determine? • Rate of the reaction • Energy absorbed or released by the reaction • Chemical names of the reactants and products • Mass of a product produced from a know mass or reactants. Answer “D” is the correct answer. This is the only answer choice that explains what we do in stoichiometry problems.

  8. 334 #3 • What is the mole ratio of CO2 to C6H12O6 in the combustion reaction: C6H12O6 + 6O2 6CO2 + 6H2O • 1:1 • 1:2 • 1:6 • 6:1 Answer “D” is the correct answer. The mole ratio is a comparison of the coefficients in front of the listed substances. Here, the coefficient for carbon dioxide is 6 and the coefficient for glucose is 1.

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