05'etc_hw_additional_answers

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					330 §05’etc – Homework
    ADDITIONAL PROBLEMS
 §5.1    Substances the Exist as Gases
 §5.2    Pressure of a Gas
 §5.3    Gas Laws
 §5.4    Ideal Gas Equation
 §5.5    Dalton’s Law of Partial Pressures
 §5.6    Kinetic-Molecular Theory of Gases
 §5.7    Deviation from Ideal Behavior

ADDITIONAL PROBLEMS                                                                          p. 167 - 169

5.92   Some commercial drain cleaners contain two components: sodium hydroxide and aluminum
         powder. When the mixture is poured down a clogged drain, the following reaction occurs:
                       2NaOH(aq) + 2Al(s) + 6H2O(l)  2NaAl(OH)4(aq) + 3H2(g)
         The heated generated in this reaction helps melt away obstructions such as grease, and the
         hydrogen gas released stirs ups the solids clogging the drain. Calculate the volume of H2 formed
         at STP if 3.12 g of Al is treated with an excess of NaOH.

        1. Stoichiometry to determine moles H2 produced.
        3.12     g Al    1        mol Al 3        mol H2 0.173      mol H2
                         26.98    g Al    2       mol Al

        2. Volume of H2 produced. *STP*
        0.173   mol H2 22.4      L      3.88      L H2
                        1        mol H2
5.93   The volume of a sample of pure HCl gas was 189 mL and 25oC and 108 mmHg. It was
         completely dissolved in about 60 mL of water and titrated with an NaOH solution: 15.7 mL of the
         NaOH solution were required to neutralize the HCl. Calculate the molarity of the NaOH solution.
      1. Number of moles of HCl:
      P=      108 mm Hg         1 atm       0.142 atm
                             760 mm Hg
      V=      189 mL          =      0.189 L
      T = 25       oC+273 =            298 K

      n=          P               V              =     0.142 atm        0.189 L-K-mol
                  R               T                   0.0821 L-atm        298 K

                                                 =   0.00110 mol HCl

      2. Molarity of NaOH:
          1 HCl + 1 NaOH  1H2O + 1NaCl
          therefore:
                               mol                           mol
           0.00110 mol HCl   1 NaOH              =   0.00110 NaOH
                             1 mol HCl

           0.00110 mole       =       0.0701 M
           0.0157 liter
5.106    A stockroom supervisor measured the contents of a partially filled 25.0-gallon acetone drum on
            a day when the temperature was 18.0oC and atmospheric pressure was 750 mmHg, and found
            that 15.4 gallons of the solvent remained. After tightly sealing the drum, an assistant dropped
            the drum while carrying it upstairs to the organic laboratory. The drum was dented and its
            internal volume was decreased to 20.4 gal. What is the total pressure inside the drum after the
            accident? The vapor pressure of the acetone at 18.0oC is 400 mmHg. (Hint: At the time the
            drum was sealed, the pressure inside the drum, which is equal to the sum of the pressures of air
            and acetone, was equal to the atmospheric pressure.)
        1A. Initial Conditions. Pressure inside container:
          PT       =   Pair     +    Pacetone
           Pair    =     PT      -        Pacetone
           Pair    =    750 mmHg             -        400 mmHg   =   350 mmHg

        1B. Initial Conditions. Volume of gas, before dent
        before: 25.0 gal        -       15.4 gal      =    9.6 gal
        after:   20.4 gal       -       15.4 gal      =    5.0 gal

        2. Pressure after (amount of air before = after; Pacetone is the same w/o DT)
           P1          V1     =        P2             V2


           P2      =      P1                     V1
                                     V2


           P2      =     350 mmHg          9.6 gal     =    672 mmHg
                                 5.0 gal
                  = pressue due to air. Pressue due to acetone is still 400 mmHg.

           PT      =    672 mmHg             +        400 mmHg
           PT      =   1072 mmHg
5.107   Lithium hydride reacts with water as follows:
                                 LiH(s) + H2O(l)  LiOH(aq) + H2(g)
          During World War II, U.S. pilots carried LiH tablets. In the event of a crash landing at sea, the
          LiH would react with the seawater and fill their life belts and lifeboats with hydrogen gas. How
          many grams of LiH are needed to fill a 4.1-L life belt at 0.97 atm and 12oC?
        n=                  ? mol
        R=             0.0821 L-atm/K-mol
        T=                 12 oC+273                  285 K

        PV = nRT                   n=                  P                         V
                                                        R                         T

                                    n=                0.97 atm                 4.1 L
                                                   0.0821 L-atm/K-mol         285 K

                                    n=                0.17 mol H2

        2. Convert moles of H2 to grams LiH:
                0.17 mol H2              1 mol LiH                  7.949 g LiH       1.4 g LiH
                                           1 mol H2                     1 mol LiH
5.109.    Nickel forms a gaseous compound of the formula Ni(CO)x. What is the value of x given the
             fact that under the same conditions of temperature and pressure methane (CH4) effuses 3.3
             times faster than the compound?

        rCH4                                MNi(CO)x
                              =                      )
       rNi(CO)x                             MCH4


      MCH4x              rCH4              MNi(CO)X
                                      =
                  rNi(CO)x

                          2
        MCH4          r       CH4           MNi(CO)x
                      2               =
                  r       Ni(CO)x

                                  2
       16.04          (3.3)                 MNi(CO)x
                                      =
                              1

           175 g/mol                  = MNi(CO)x


      Subtracting mass of Ni in Ni(CO)x :
        Ni:     58.69  116.31         g/mol

      Each CO:                        = 28.01           g/mol

                                      = 4.152446  4           Ni(CO)4

       (x = 4)

				
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