Stoichiometric Calculations -   Limiting Reagents and Percentage Yield "If one reactant is entirely used up before any of the other reactants, then that reactant limits the maximum yield of the product." Problems of this type are done in exactly the same way as the previous examples, except that a decision is made before the ratio comparison is done. The decision that is made is "What reactant is there the least of?" Example Problem #1 Methane, CH4, burns in oxygen to give carbon dioxide and water according to the following equation:                           CH4  + 2 O2 ------>  CO2 + 2 H2O In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel vessel.  Find the limiting reactant, if any, and calculate the theoretical yield, (in moles) of water. Solution: In any limiting reactant question, the decision can be stated in two ways.  Do it once to get an answer, then do it again the second way to get a confirmation. According to the equation:    1 mol CH4 = 2 mol O2 If we use up all the methane then:                               1 mol CH4 = 2 mol O2                                  0.25 mol          x                      x = 0.50 mol of O2 would be needed. We have 1.25 mol of O2 on hand. Therefore we have 0.75 mol of O2 in excess of what we need. If the oxygen in is excess, then the methane is the limiting reactant. Confirmation:  If we use up all the oxygen then                             1 mol CH4 = 2 mol O2                                     x             1.25 mol          x = 0.625 mol of methane. We don't have 0.625 moles of methane. We have only 0.25 moles.  Therefore the methane will be used up before all the oxygen is.  Again the methane is the limiting reactant. We now use the limiting reactant to make the mole comparison across the bridge to find the amount of water produced. 1 mol CH4 = 2 H2O                               0.25 mol         x             x = 0.50 mol of H2O  would be produced. Finish off with a statement:  When 0.25 mole of methane and 1.25 mole of oxygen are mixed and reacted according to the equation, the methane is the limiting reactant and the maximum yield of water will be 0.50 moles. Example Problem #2 Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl.  In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed.  Which is the limiting reactant?  What is the maximum yield of CCl4 in moles and in grams? Solution: Start with the equation:        CHCl3  +  Cl2 ------->  CCl4  +  HCl Did you check to see if it was balanced? Calculate the molecular masses of the species needed in the problem. CHCl3 = 1 C = 1(12.01)  =    12.01                           Cl2 = 2 (35.45) = 70.90 g/mol                1 H =  1(1.01)  =       1.01                           H2O = 2 H = 2  (1.01)  =   2.02                3 Cl =  3(35.45) = 106.35                                       1 O = 1 (16.00) = 16.00                                             119.37 g/mol                                                         18.02 g/mol Then calculate the moles of each of the reactants to be used. moles of CHCl3 =   g   =    25.00 g        =  0.21 moles of CHCl3 are present.                              mm   119.37 g/mol moles of Cl2 =    g   =     25.00 g       =  0.35 moles of chlorine are present.                         mm        70.90 g/mol Decision time.  Which of the two reactants do you have the least of? From the balanced equation you can see that the chloroform and chlorine reactant in a one to one ratio.  If we use all the chloroform then we get the following equation.                       1 CHCl3 =   1 Cl2                       0.21 mol           x                     x = 0.21 moles of chlorine are needed. We need 0.21 moles of chlorine.  We have 0.35 moles of chlorine.  Therefore chlorine is in excess.  The chloroform must be the limiting reactant. Confirmation:  IF we use all the chlorine then:                        1 CHCl3 =   1 Cl2                            x             0.35 mol              x = 0.35 moles of chloroform are needed. If we use all the chlorine then we need 0.35 moles of chloroform.  We have only 0.21 moles of chloroform.  It is the reactant that we will run out of first.  Therefore it is the limiting reactant. Use the limiting reactant to cross the ratio bridge and find the number of moles of water made.                              1 CHCl3  =  2 H2O                              0.21 mol         x            x = 0.42 moles of H2O will be made. Calculate the grams of water produced.    grams = moles * molecular mass                                                                          = 0.42 mol * 18.02 g/mol                                                                          = 7.57 grams of water Finish off with a statement:   When 25 grams of each reactant are mixed according to the equation, the chloroform is the limiting reagent and the maximum yield of water will be 0.42 moles or 7.57 grams. Example Problem #3 Aluminum chloride, AlCl3, can be made by the reaction of aluminum with chlorine according to the following equation:                                       2 Al  + 3 Cl2  ------> 2  AlCl3 What is the limiting reactant if 20.0 grams of Al and 30.0 grams of Cl2 are used, and how much AlCl3 can theoretically form? Have you checked to make sure the equation is balanced correctly? Find the molecular masses of all species involved. Al = 26.98 g/mol      Cl2 =  70.90 g/mol      AlCl3 =  133.33 g/mol Convert the grams into moles. moles of Al = g/mm = 20.00 g/26.98 g/mol = 0.74 moles of aluminum on hand. moles of Cl2 = g/mm = 30.00 g/70.90 g/mol = 0.42 moles of chlorine on hand. Decision time:  Which is the limiting reagent? IF we use all aluminum then:                                2 Al       =   3  Cl2                             0.74 mol          x          x = 1.11 moles of chlorine are needed. We don't have 1.11 moles of chlorine.  We have 0.42 moles of chlorine. Therefore we will run out of chlorine first. It is the limiting reactant. Confirmation: If we use all the chlorine then:                                 2 Al       =   3  Cl2                                    x            0.42 mol        x = 0.28 moles of aluminum are needed. We have 0.74 moles of aluminum, therefore it is in excess. If it is in excess then the chlorine is the limiting reactant. Use the limiting reactant to cross the ratio bridge and find the moles of AlCl3 that will be produced.                                       3 Cl2   =        2 AlCl3                                   0.42 mol             x                   x = 0.28 moles of AlCl3 are produced Grams of aluminum chloride are found with g = n * mm = 0.28 mol * 133.33 g/mol  = 37.33 g Finishing statement:  When 20.0 grams of aluminum and 30.0 grams of chlorine are reacted according to the above equation, the chlorine is the limiting reactant and the maximum yield of aluminum chloride is 0.28 moles or 37.33 grams. Stoichiometry Limiting Reagents and % Yield Calculations Worksheet Stoichiometry Impure Samples and Percentage Purity Worksheet Stoichiometry Percentage Yield Problems Worksheet