ch3 - 3 3.1 3.2 2 KClO3 3.3 3.4 3 A2 + 2 B (a) (b) (c) (d)...

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3 Formulas, Equations, and Moles 3.1 2 KClO 3 2 KCl + 3 O 2 3.2 (a) C 6 H 12 O 6 2 C 2 H 6 O + 2 CO 2 (b) 4 Fe + 3 O 2 2 Fe 2 O 3 (c) 4 NH 3 + Cl 2 N 2 H 4 + 2 NH 4 Cl 3.3 3 A 2 + 2 B 2 BA 3 3.4 (a) Fe 2 O 3 : 2(55.85) + 3(16.00) = 159.7 amu (b) H 2 SO 4 : 2(1.01) + 1(32.07) + 4(16.00) = 98.1 amu (c) C 6 H 8 O 7 : 6(12.01) + 8(1.01) + 7(16.00) = 192.1 amu (d) C 16 H 18 N 2 O 4 S: 16(12.01) + 18(1.01) + 2(14.01) + 4(16.00) + 1(32.07) = 334.4 amu 3.5 Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g) 0.500 mol CO mol 1.50 = O Fe mol 1 CO mol 3 x O Fe 3 2 3 2 3.6 C 5 H 11 NO 2 S: 5(12.01) + 11(1.01) + 1(14.01) + 2(16.00) + 1(32.07) = 149.24 amu 3.7 C 9 H 8 O 4 , 180.2 amu; 500 mg = 500 x 10 3 g = 0.500 g 0.500 g x aspirin mol 10 x 2.77 = g 180.2 mol 1 3 _ 2.77 x 10 3 mol x molecules aspirin 10 x 1.67 = mol 1 molecules 10 x 6.02 21 23 3.8 salicylic acid, C 7 H 6 O 3 , 138.1 amu; acetic anhydride, C 4 H 6 O 3 , 102.1 amu aspirin, C 9 H 8 O 4 , 180.2 amu; acetic acid, C 2 H 4 O 2 , 60.1 amu 4.50 g C 7 H 6 O 3 x O H C mol 1 O H C g 102.1 x O H C mol 1 O H C mol 1 x O H C g 138.1 O H C mol 1 3 6 4 3 6 4 3 6 7 3 6 4 3 6 7 3 6 7 = 3.33 g C 4 H 6 O 3 4.50 g C 7 H 6 O 3 x O H C mol 1 O H C g 180.2 x O H C mol 1 O H C mol 1 x O H C g 138.1 O H C mol 1 4 8 9 4 8 9 3 6 7 4 8 9 3 6 7 3 6 7 = 5.87 g C 9 H 8 O 4 4.50 g C 7 H 6 O 3 x O H C mol 1 O H C g 60.1 x O H C mol 1 O H C mol 1 x O H C g 138.1 O H C mol 1 2 4 2 2 4 2 3 6 7 2 4 2 3 6 7 3 6 7 = 1.96 g C 2 H 4 O 2 33
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3.9 C 2 H 4 , 28.1 amu; C 2 H 6 O, 46.1 amu 4.6 g O H C mol 1 O H C g 46.1 x H C mol 1 O H C mol 1 x H C g 28.1 H C mol 1 x H C 6 2 6 2 4 2 6 2 4 2 4 2 4 2 = 7.5 g C 2 H 6 O (theoretical yield) % 63 = % 100 x g 7.5 g 4.7 = % 100 x yield l Theoretica yield Actual = yield Percent 3.10 CH 4 , 16.04 amu; CH 2 Cl 2 , 84.93 amu; 1.85 kg = 1850 g 1850 g CH 4 x Cl CH mol 1 Cl CH g 84.93 x CH mol 1 Cl CH mol 1 x CH g 16.04 CH mol 1 2 2 2 2 4 2 2 4 4 = 9800 g CH 2 Cl 2 (theoretical yield) Actual yield = (9800 g)(0.431) = 4220 g CH 2 Cl 2 3.11 Li 2 O, 29.9 amu: 65 kg = 65,000 g; H 2 O, 18.0 amu: 80.0 kg = 80,000 g 65,000 g Li 2 O x O Li g 29.9 O Li mol 1 2 2 = 2.17 x 10 3 mol Li 2 O 80,000 g H 2 O x O H g 18.0 O H mol 1 2 2 = 4.44 x 10 3 mol H 2 O The reaction stoichiometry between Li 2 O and H 2 O is one to one. There are twice as many moles of H 2 O as there are moles of Li 2 O. Therefore, Li 2 O is the limiting reactant. (4.44 x 10 3 mol 2.17 x 10 3 mol) = 2.27 x 10 3 mol H 2 O remaining 2.27 x 10 3 mol H 2 O x O H mol 1 O H g 18.0 2 2 = 40,860 g H 2 O = 40.9 kg = 41 kg H 2 O 3.12 LiOH, 23.9 amu; CO 2 , 44.0 amu CO g 921 = CO mol 1 CO g 44.0 x LiOH mol 1 CO mol 1 x LiOH g 23.9 LiOH mol 1 x LiOH g 500.0 2 2 2 2 3.13 (a) A + B 2 AB 2 There is a 1:1 stoichiometry between the two reactants. A is the limiting reactant because there are fewer reactant A's than there are reactant B 2 's. (b)
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This note was uploaded on 08/27/2008 for the course CHEM 1280 taught by Professor Schmidt during the Fall '07 term at Toledo.

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ch3 - 3 3.1 3.2 2 KClO3 3.3 3.4 3 A2 + 2 B (a) (b) (c) (d)...

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