Stoichiometry Workshop - Solutions

Chemistry 11 Fall 2010 Stoichiometry Workshop - Solutions 1. Fe2O3 (s) + 2 Al(s)  Al2O3 (s) + 2 Fe() 150 g Al  1 mol Al 2 mol Fe 55.85 g Fe    311 g Fe 26.98 g Al 2 mol Al 1 mol Fe 250 g Fe2O 3    1 mol Fe2O 3 2 mol Fe 55.85 g Fe    175 g Fe 159.70 g Fe2O 3 1 mol Fe2O 3 1 mol Fe Fe2O3 is the limiting reagent, and 175 g Fe is produced from this reaction. 2. Consider 100.0 g of vitamin A: 100.00 g vitamin A 83.86 g C  1 mol vitamin A  0.3492 mol vitamin A 286.4 g vitamin A 1 mol C  6.982 mol C 12.011 g C 6.982 mol C 19.99 mol C 20 mol C   0.3492 mol vitamin A 1 mol vitamin A 1 mol vitamin A  10.56 g H   1 mol H 10.48 mol H 1.0079 g H 10.48 mol H 30.01 mol H 30 mol H   0.3492 mol vitamin A 1 mol vitamin A 1 mol vitamin A (100 – 83.86– 10.56) g E 15.98 g E 1 mol vitamin A 15.98 g E    0.3492 mol vitamin A 1 mol vitamin A 1 mol E 1 mol E  The molecular weight of E is 15.98 g/mol, which suggests that E is oxygen (O) and the molecular formula of vitamin A is C20H30O.  3. a. By conservation of mass: 24.96 g CO 2 12.27 g H 2O  10.00 g C 5H12O  27.23 g O 2  1 mol CO2 1 mol C 12.01 g C b. 24.96 g CO2    0.5671 mol C  6.811 g C 44.01 g CO2 1 mol CO2 1 mol C  1 mol H2O 2 mol H 1.008 g H 12.27 g H2O    1.362 mol H   1.373 g H 18.02 g H2O 1 mol H2O 1 mol H mass of O = 10.00 g MTBE  6.811 g C  1.373 g H 1.82 g O 1.82 g O     1 mol O  0.1135 mol O 16.00 g O Chemistry 11 Fall 2010 Dividing each mole value by 0.1135 yields a mole ratio of 5 C : 12 H : 1 O. The empirical formula is thus (C5H12O)n, which has a formula weight of 88.15 amu. c. The MTBE vapor has a mass about twice the mass of an equal volume of carbon dioxide gas at the same temperature and pressure. This implies that the molecular mass of MTBE is 2 times the molecular mass of carbon dioxide gas, which is 44.01 amu. Thus, the molecular mass of the sample must be ≈ 88 amu. The mass of the empirical formula unit C5H12O is 88.15 amu. Thus, n = 1, and the molecular formula of MTBE must be C5H12O. d. 2 C5H12O(l) + 15 O2 (g)  10 CO2 (g) + 12 H2O(g) 4. a. CaCO3 (s) + 2 HCl(aq)  CaCl2 (aq) + CO2 (g) + H2O(g) b. MgCO3 (s) + 2 HCl(aq)  MgCl2 (aq) + CO2 (g) + H2O(g) c. 1.000 g of lime contains 0.800 CaCO3 and 0.200 MgCO3. 0.800 g CaCO3  1 mol CaCO3 1 mol CO2 44.01 g CO2    0.352 g CO 2 100.05 g CaCO3 1 mol CaCO3 1 mol CO2 0.200 g MgCO3    total CO2 = (0.352 g) + (0.104 g) = 0.456 g CO2 d. 1.000 g of lime contains 0.800 CaCO3 and 0.200 MgCO3. 0.800 g CaCO3  1 mol CaCO3 2 mol HCl 36.46 g HCl    0.583 g HCl 100.05 g CaCO3 1 mol CaCO3 1 mol HCl 0.200 g MgCO3    e. 1.000 g mixture = (x g CaCO3) + (y g MgCO3) 1 mol CaCO3 1 mol CO2 44.01 g CO2    0.440x g CO 2 from CaCO3 100.05 g CaCO3 1 mol CaCO3 1 mol CO2 y g MgCO 3   1 mol MgCO3 2 mol HCl 36.46 g HCl    0.173 g HCl 84.32 g MgCO3 1 mol CaCO3 1 mol HCl total HCl = (0.583 g) + (0.173 g) = 0.756 g HCl x g CaCO 3   1 mol MgCO3 1 mol CO2 44.01 g CO2    0.104 g CO 2 84.32 g MgCO3 1 mol MgCO3 1 mol CO2 1 mol MgCO3 1 mol CO2 44.01 g CO2    0.522y g CO 2 from MgCO3 84.32 g MgCO3 1 mol MgCO3 1 mol CO2 0.500 g CO2 = (0.440x g CO2 from CaCO3) + (0.522y g CO2 from MgCO3) 0.500 g CO2 = (0.440x g CO2) + (0.522  (1.000–x) g CO2) 0.500 g CO2 = 0.522 g CO2 – 0.082x g CO2 x (0.522 – 0.500) g CO2  0.268 0.082 g CO2 The mixture contains 0.268 g CaCO3 and 0.732 g MgCO3. 