Page 157 - Theory and Problems of BEGINNING CHEMISTRY
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146                                  STOICHIOMETRY                               [CHAP. 10


                     These quantities are entered in line 2 also.
                                                           4Al    + 3O 2   −→ 2Al 2 O 3
                                       Present initially:  4.60 mol  4.20 mol  0.00 mol
                                       Change due to reaction:  4.60 mol  3.45 mol  2.30 mol
                                       Present finally:
                     The final line is calculated by subtraction for the reactants and addition for the product.
                                                           4Al    + 3O 2   −→ 2Al 2 O 3
                                       Present initially:  4.60 mol  4.20 mol  0.00 mol
                                       Change due to reaction:  4.60 mol  3.45 mol  2.30 mol
                                       Present finally:     0.00 mol  0.75 mol  2.30 mol

               EXAMPLE 10.9. How many moles of PbI 2 can be prepared by the reaction of 0.128 mol of Pb(NO 3 ) 2 and 0.206 mol NaI?
               Ans.  The balanced equation is
                                                Pb(NO 3 ) 2 + 2 NaI −→ PbI 2 + 2 NaNO 3
                     The limiting quantity is determined:
                               0.128 mol Pb(NO 3 ) 2                  0.206 mol NaI
                                              = 0.128 mol Pb(NO 3 ) 2            = 0.103 mol NaI
                                      1                                    2
                     NaI is in limiting quantity.

                                                         1 mol PbI 2
                                             0.206 mol NaI         = 0.103 mol PbI 2
                                                         2 mol NaI
                     Note especially that the number of moles of NaI exceeds the number of moles of Pb(NO 3 ) 2 present, but the NaI is
                     still in limiting quantity.
               EXAMPLE 10.10. How many grams of Ca(ClO 4 ) 2 can be prepared by treatment of 12.0 g CaO with 102 g HClO 4 ?How
               many grams of excess reactant remains after the reaction?
               Ans.  This problem gives the quantities of the two reactants in grams; we must first change them to moles:

                                                         1 mol CaO
                                               12.0 g CaO          = 0.214 mol CaO
                                                         56.0 g CaO

                                                        1 mol HClO 4
                                            102.0 g HClO 4         = 1.02 mol HClO 4
                                                        100 g HClO 4
                     Now the problem can be done as in Example 10.8. All quantities are in moles.
                     The balanced equation is
                                                         CaO + 2 HClO 4 −→ Ca(ClO 4 ) 2 + H 2 O
                                    Present initially:   0.214  1.02      0.000
                                    Change due to reaction:  0.214  0.428  0.214
                                    Present finally:      0.000  0.59      0.214


                                                     239gCa(ClO 4 ) 2
                                    0.214 mol Ca(ClO 4 ) 2         = 51.1gCa(ClO 4 ) 2 produced
                                                     1 mol Ca(ClO 4 ) 2
                                                        100 g HClO 4

                                          0.59 mol HClO 4          = 59 g HClO
                                                                             4
                                                        1 mol HClO 4
                   If the quantities of both reactants are in exactly the correct ratio for the balanced chemical equation, then
               either reactant may be used to calculate the quantity of product produced. (If on a quiz or examination it is
               obvious that they are in the correct ratio, you should state that they are, so that your instructor will understand
               that you recognize the problem to be a limiting-quantities type problem.)
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