Organotin compounds play a significantrole in diverse industrial applications. T

Question

Organotin compounds play a significantrole in diverse industrial applications. They have been used asplastic stabilizers and as pesticides or fungicides.
One method used to prepare simple tetraalkylstannanes isthe controlled direct reaction of liquid tin(IV) chloride withhighly reactive trialkylaluminum compounds, such as liquidtriethylaluminum(Al(C2H5)3).

In one experiment, 0.180 L of SnCl4 (d = 2.226 g/mL)was treated with 0.317 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)?
If 0.256 L of tetraethylstannane (d = 1.187 g/mL) wereactually isolated in this experiment, what was the percentyield?

Explanation / Answer

3SnCl4 + 4Al(C2H5)3 ----> 3Sn(C2H5)4 +4AlCl3 Molar mass of SnCl4 = 119 + 4 * 35.5 = 261 g Molar mass of Al(C2H5)3 is 27 + 3( 2*12 + 5 * 1 ) = 114 g Molar mass of Sn(C2H5)4 is = 119 + 4 ( 2 * 12 + 5 * 1 ) = 235 g We know that density , d = mass / Volume So , mass , m = d * Volume m SnCl4   = 2.226 g / mL * 180 mL             = 400.68 g m Al(C2H5)3=0.835 g / mL * 317 mL                   = 264.695g 3 * 261 (= 783g) of SnCl4 reacts with 4 * 114(=456g) of Al(C2H5)3 400.68 g of SnCl4 reacts with X g of Al(C2H5)3 X = ( 456 * 400.68 ) / 783 = 233.346 g ofAl(C2H5)3 So , Al(C2H5)3   isthe excess reactant. 783 g of SnCl4 reacts produces 3* 235 g ofSn(C2H5)4 400.68 g of SnCl4 reacts produces Y g ofSn(C2H5)4 Y = ( 3 * 235 * 400.68 ) / 783 = 360.765 g ofSn(C2H5)4      --- this is the theoretical yield Actual yield = density * Volume                   = 1.187 g / mL * 256 mL                   = 303.872 g So , % yield = ( actual / theoretical yield ) * 100                   = 84.23 % m Al(C2H5)3=0.835 g / mL * 317 mL                   = 264.695g 3 * 261 (= 783g) of SnCl4 reacts with 4 * 114(=456g) of Al(C2H5)3 400.68 g of SnCl4 reacts with X g of Al(C2H5)3 X = ( 456 * 400.68 ) / 783 = 233.346 g ofAl(C2H5)3 So , Al(C2H5)3   isthe excess reactant. 783 g of SnCl4 reacts produces 3* 235 g ofSn(C2H5)4 400.68 g of SnCl4 reacts produces Y g ofSn(C2H5)4 Y = ( 3 * 235 * 400.68 ) / 783 = 360.765 g ofSn(C2H5)4      --- this is the theoretical yield Actual yield = density * Volume                   = 1.187 g / mL * 256 mL                   = 303.872 g So , % yield = ( actual / theoretical yield ) * 100                   = 84.23 % 3 * 261 (= 783g) of SnCl4 reacts with 4 * 114(=456g) of Al(C2H5)3 400.68 g of SnCl4 reacts with X g of Al(C2H5)3 X = ( 456 * 400.68 ) / 783 = 233.346 g ofAl(C2H5)3 So , Al(C2H5)3   isthe excess reactant. 783 g of SnCl4 reacts produces 3* 235 g ofSn(C2H5)4 400.68 g of SnCl4 reacts produces Y g ofSn(C2H5)4 Y = ( 3 * 235 * 400.68 ) / 783 = 360.765 g ofSn(C2H5)4      --- this is the theoretical yield Actual yield = density * Volume                   = 1.187 g / mL * 256 mL                   = 303.872 g So , % yield = ( actual / theoretical yield ) * 100                   = 84.23 % 400.68 g of SnCl4 reacts produces Y g ofSn(C2H5)4 Y = ( 3 * 235 * 400.68 ) / 783 = 360.765 g ofSn(C2H5)4      --- this is the theoretical yield Actual yield = density * Volume                   = 1.187 g / mL * 256 mL                   = 303.872 g So , % yield = ( actual / theoretical yield ) * 100                   = 84.23 % Y = ( 3 * 235 * 400.68 ) / 783 = 360.765 g ofSn(C2H5)4      --- this is the theoretical yield Actual yield = density * Volume                   = 1.187 g / mL * 256 mL                   = 303.872 g So , % yield = ( actual / theoretical yield ) * 100                   = 84.23 %

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