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Chem Stoich problem

  1. Sep 18, 2007 #1
    Organotin compounds play a significant role in diverse industrial applications. They have been used as plastic stabilizers and as pesticides or fungicides.
    One method used to prepare simple tetraalkylstannanes is the controlled direct reaction of liquid tin(IV) chloride with highly reactive trialkylaluminum compounds, such as liquid triethylaluminum (Al(C2H5)3).

    3SnCl4 + 4Al(C2H5)3 --> 3Sn(C2H5)4 + 4AlCl3

    In one experiment, 0.230 L of SnCl4 (d = 2.226 g/mL) was treated with 0.497 L of triethylaluminum (Al(C2H5)3); d = 0.835 g/mL).
    What is the theoretical yield in this experiment (mass of tetraethylstannane, Sn(C2H5)4)?

    My attempt:

    #mol SnCl4 = .230 L x 2226g/L x 1mol/260.6g
    = 1.96537428 mol SnCl4

    #mol Al(C2H5)3 = .497L x 835g/L x 1mol/114.16g
    = 3.645204975 mol Al(C2H5)3

    3SnCl4 mol : 4Al(C2H5)3 mol

    1.97mol SnCl4 x 4mol Al(C2H5)3 / 4 mol SnCl4
    = 2.66 mol Al(C2H5)3

    Sn Cl4 is limiting

    #g Sn(C2H5)4 = 7.86146712 mol SnCl4 x (3 mol SnCl4 / 3mol Sn(C2H5)4) x 234.94g / mol
    = 1846.980133g

    As far as I know I did all the steps correctly, so I'm not sure where I went went. I have checked my calculations over but I'm still getting a wrong answer.

  2. jcsd
  3. Sep 18, 2007 #2
    Disclaimer: I'm just a student too. I could be completly off.
    I also used this periodic table: http://highered.mcgraw-hill.com/ibis/cds/classware/PeriodicTable.swf

    First, I figured out what the molar mass of each compound was.


    1 Sn: 118.69
    4 Cl: 35.453*4=141.812
    SnCl_4 = 260.5036 (I don't apply sig figs until I come to the answer)


    1 Al: 26.98
    6 C: 72.066
    15 H: 15.12
    Al(C_2H_5)_3 = 88.186

    Then, I took D=M/V to figure out how many grams of each element I was working with, by multiplying both sides of the equation by V, since we are given D and V. After converting L to mL, this is what I came up with

    230 mL of SnCl_4 * 2.226 g/mL = 511.98 g SnCl_4 (mL cancel)
    497 mL Al(C_2H_5)_3 * 0.835 g/mL = 414.995 g Sn(C_2H_5)_3

    Then I set up my stoichemetry, and plugged the numbers:

    511.98 g SnCl_4 * 1 mol SnCl_4/260.5036 g * 4 mol Al(C_2H_5)_3/3 mol SnCl_4 * 88.186 g Al(C_2H_5)_3/1 mol Al(C_2H_5)_3 = 231 g Al(C_2H_5)_3 (after sig figging it)

    Honestly, just took for granted that SnCl_4 was the limiting reagent, so I converted it into how many g of Al(C_2H_5)_3 you'd find. Hope that helps!
  4. Sep 18, 2007 #3

    1 Al: 26.98
    6 C: 72.066
    15 H: 15.12
    Al(C_2H_5)_3 = 88.186

    how did you get 88.186?
    those 3 values add up to be 114.166

  5. Sep 19, 2007 #4
    Miscalculation :). Like I said, student here as well.
  6. Sep 19, 2007 #5
    ok, thanks for your help anyways
    I'll post the solution if I can figure it out
  7. Sep 19, 2007 #6
    1.96537428 mol SnCl4

    not 7.86146712 SnCl4 for final equation

    = 4.62×10^2 g

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