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3 Stoichiometry Solutions to Exercises 3.104 (a) 1.5x10⁵ g C₉H₈O₄ X C₇H₆O₃ 138.1g C₇H₆O₃ 180.2 1 mol 1 mol C₇H₆O₃ = 1.1496 g = 1.1 kg C₇H₆O₃ (b) If only 80 percent of the acid reacts, then we need 1/0.80 = 1.25 times as much to obtain the same mass of product: 1.25 X 1.15 10² kg = 1.4 X 10² kg C₇H₆O₃ (c) Calculate the number of moles of each reactant: g C₇H₆O₃ 1 mol C₇H₆O₃ = = 1.34x10³ mol C₇H₆O₃ g C₄H₆O₃ 1 mol C₄H₆O₃ C₄H₆O₃ = 1.224x = 1.22x10³ mol C₄H₆O₃ We see that C₄H₆O₃ limits, because equal numbers of moles of the two reactants are consumed in the reaction. 1.224x10³ mol 1 1 mol mol C₇H₆O₃ 180.2 C₉H₈O₄ = = (d) percent Integrative Exercises 3.105 Plan. Volume cube density mass CaCO₃ moles CaCO₃ moles atoms Solve. (2.005)³ (2.54)³ 1in³ cm³ 2.71 g CaCO₃ 100.1 1 mol CaCO₃ CaCO₃ X 1 mol CaCO₃ atoms = Datoms 1mol 3.106 (a) Plan. volume of Ag cube density mass of Ag -> mol Ag Ag atoms Solve. (1.000)³ cm³ Agx 10.5g Ag Ag 107.87 1 mol Ag Ag 6.022 atoms = = Ag atoms (b) 1.000 cm³ cube volume, 74% is occupied by Ag atoms 0.74 cm³ = volume of 5.86 10²² Ag atoms 0.7400 cm³ = 1.2624 = / Ag atom 5.8618x10²² Ag atoms Since atomic dimensions are usually given in Å, we will show this conversion. cm³ 1ų m³ = 12.62 = 13 / Ag atom 74