QUIZ OCTOBER 9-10 ANSWERS

  1. Completing reactions
    1. Electrolysis of water.
      2 H2O 2 H2 + O2

    2. From lab #1:
      FeCl3 + 3 NH4OH Fe(OH)3 + 3NH4Cl

    3. An extrapolation (going from what you have known to figuring out something new):
      Fe(OH)3 + 3 HNO3 Fe(NO3)3 + 3 H2O

  2. Mary Lou's work:
    The first thing to tell you is that the problem was not the same for both sections. On Monday she took 820 mg of Li2SO4 dissolved in 250 ml water plus a trace of HCl. After heating she added 100 ml of 0.1 M Ba(NO3)2. How much precipitate was expected to be obtained? On Tuesday, only 50 ml of 0.1 M Ba(NO3)2 was added.

    1. The limiting reactant on Monday was the Li2SO4, while on Tuesday it was the Ba(NO3)2. Let's see how this happens.

      820 mg of Li2SO4 = 820/109.9 = 7.46 millimoles.

      100 ml of 0.1 M Ba(NO3)2 = 10 millimoles (ml x moles = millimoles), which more than enough to react with ALL of the 7.46 millimoles of Li2SO4.

      50 ml of 0.1 M Ba(NO3)2 = 5 millimoles, but this is not enough to react with all of the Li2SO4.

      THUS on Monday 7.46 millimoles of BaSO4 should have been formed, and on Tuesday 5.00 millimoles should have been formed.

      THEREFORE: 7.46 x 233.4 = 1,741 mg = 1.741 gram on Monday
      and 5.00 x 233.4 = 1,167 mg = 1.167 gram on Tuesday.

    2. The flow diagram is exactly the same as on one of the handouts you should have had with you - the handout on making flow charts.