Stoichiometry Test Study Guide
2H2O (l)  2H2(g) + O2 (g) Start 5.8 mol H2O, how many of each product? 5.8 mol H2O x 2 mol H2 = 5.8 mol H2 2 mol H2O 5.8 mol H2O x 1 mol O2 = 2.9 mol O2 2 mol H2O C3H8 (propane) C3H8 (g) + 5 O2  CO2 (g) + 4H2O (l) 96.1 g ? g What mass of oxygen will be required to completely react 96.1g C3H8? 96.1 g C3H8 x 1 mol C3H8 x 5 mol O2 x 32.0 g O2= 349 g O2 44.0 g C3H8 1mol C3H8 1mol O2 Grams Moles Moles Grams A B B A Molar Mole Molar mass A ratio mass B Lithium hydroxide to remove exhaled carbon dioxide. The products are solid lithium carbonate and water. What mass of CO2 can 1.00 x 103 g of Lithium hydroxide absorb? What volume of CO2 is absorbed @ STP. 2LiOH (s) + CO2 (g)  Li2CO2 (s) +H2O (g) 1.00 x 103 g ? g 1.00 x 103 g LiOH x 1 mol LiOH x 1 mol CO2 X 22.4 L CO2 = 467 L CO2 22.40 g LiOH 2 mol LiOH 1 mol CO2 Limiting Reactant • Reactant that creates the least amount of product 2NH3 (g) + 3 CUO(s)  N2 (g) + 3 Cu (s) + 3H2O (g) 1) How many grams of N2 are formed when 18.1 g NH3 reacts with 90.4 g CuO? 2) Identify the Limiting Reactant (LR) and excess reactant (ER/XS) 3) How much ER is left over? 1) 18.1 g NH3 x 1 mol NH3 x 1 mol N2 x 28.0 g N2 = 14.9 g N2 17.0 g NH3 2 mol NH3 1 mol N2 90.4 g CuO x 1 mol CuO x 1 mol N2 x 28.0 g N2 = 10.6 g N2 79.6 g CuO 3 mol CuO 1 mol N2 2) LR=CuO ER=NH3 3) Mass of LR convert mass of ER 90.4 g CuO x 1 mol CuO x 2 mol NH3 x 17.0 g NH3 = 12.9 g NH3 79.6 g CuO 3 mol CuO 1 mol NH3 18.1 g NH3 (started) -­‐12.9 g NH3 (used) 5.2 g NH3 (in excess)-­‐ Amount of ER in "excess" Percent Yield % Yield = Actual Yield x 100 Theoretical Yield Actual Yield : amount of product produced Theoretical Yield: Amount of product formed by the limiting reactant Example 1) CO (g) + 2H2 (g) -­‐-­‐> CH3OH (l) 68.5 kg 8.60 kg A) Calculate the theoretical yield of the product. B) If 3.57 x 104 g of product is actually produced, what is % yield? A) 68.5 kg CO x 1 x 103 g CO x 1 mol CO x 1 mol CH3OH x 32.0 g CH3OH = 7.83 x 104 g CH3OH 1 kg CO 28.0g CO 1 mol CO 1 mol CH3OH 8.60 kg H2 x 1 x 103 g H2 x 1 mol H2 x 1 mol CH3OH x 32.0 g CH3OH = 6.88 x 104 g CH3OH 1 kg H2 2.00g H2 2 mol H2 1 mol CH3OH 4 LR: H2 Theoretical Yield: 6.88 x 10 g CH3OH ER: CO B) % Yield= 3.57 x 104 g x 100 = 51.9 % 6.88 x 104 g Example 2) Titanium IV Chloride ( 6.7 x 103 g ) + Oxygen gas ( 2.45 x 13 g )-­‐-­‐> Titanium IV Oxide + Chlorine gas TiCl4 (s) + O2 (g) -­‐-­‐> TiO2 (s) + 2Cl2 A) Calculate theoretical yield. B) Calculate the actual yield of titanium IV oxide % yield is 75% A) 6.71 x 103 g TICl4 x 1 mol TICl4 x 1 mol TiO2 x 79.9 g TiO2 = 2.82 x 103 g TiO2 189.9 gTiCl4 1 mol TiCl4 1 mol TiO2 2.45 x 103 g O2 x 1 mol O2 x 1 mol TiO2 x 79.9 g TiO2 = 6.12 x 103 g TiO2 32.0 g O2 1 mol O2 1 mol TiO2 LR-­‐ TiCl4 ER-­‐ O2 Theoretical yield: 2.82 x 103 g TiO2 B) 0.75= x = 2.12 x 103 (actual yield) 2.82 x 103