Calculate Theoretical Yield of Sn(C2H5)4 from SnCl4 & Al(C2H5)3

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Discussion Overview

The discussion revolves around calculating the theoretical yield of tetraethylstannane (Sn(C2H5)4) from the reaction of tin(IV) chloride (SnCl4) and triethylaluminum (Al(C2H5)3). Participants explore the stoichiometry involved in the reaction and the calculations necessary to determine the yield, including the identification of limiting reagents.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant outlines the reaction and their calculations for the theoretical yield, stating that SnCl4 is the limiting reagent.
  • Another participant provides their own calculations for the molar masses of SnCl4 and Al(C2H5)3, arriving at different values for the mass of Al(C2H5)3 based on their stoichiometric setup.
  • Several participants question the molar mass calculation of Al(C2H5)3, noting discrepancies in the values presented, specifically pointing out that the sum of the individual atomic masses does not match the claimed molar mass.
  • A later reply corrects the earlier molar mass calculation for SnCl4, indicating a misunderstanding in the final equation used for yield calculation.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the molar mass calculations for Al(C2H5)3, with some agreeing on the need for correction while others maintain their original values. The discussion remains unresolved regarding the correct theoretical yield due to these discrepancies.

Contextual Notes

There are limitations in the calculations presented, including potential miscalculations of molar masses and the final yield equation. Participants rely on different assumptions and methods for their calculations, leading to varied results.

Who May Find This Useful

This discussion may be useful for students studying organic chemistry, particularly those interested in organotin chemistry and stoichiometric calculations in chemical reactions.

Destrio
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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.

Thanks
 
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Disclaimer: I'm just a student too. I could be completely 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.

SnCl_4

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)

Al(C_2H_5)_3

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!
 
Al(C_2H_5)_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

thanks
 
Destrio said:
Al(C_2H_5)_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

thanks

Miscalculation :). Like I said, student here as well.
 
ok, thanks for your help anyways
I'll post the solution if I can figure it out
 
1.96537428 mol SnCl4

not 7.86146712 SnCl4 for final equation

= 4.62×10^2 g

solved