Thursday, November 10, 2005

My teachers ruined my break.

Here's some of the easy crap I have to do. (I mean it's really easy)

The reaction of ethane gas(C2H6) with Clhorine gas produces C2H5Cl as its main product (along with HCl). In addition, the reaction produces C2H4Cl2, C2H3Cl3, and others. Naturally the production of these minor products reduces the yeild of the main product. Calculate the percent yeild of C2H5Cl if the reaction of 300. g of ethane with 650. g of Chlorine produced 490. g of C2H5Cl.

_____________________
- 490g C2H5Cl
-300g C2H6
-650g Cl
-% yelid of C2H5Cl?
-
-____________________

C2H6(g) + 2Cl(g) ------------> C2H5Cl + HCl

300 g C2H6 X
1 mol C2H6/ 30g C2H6 = 10 mol C2H6

650g Cl X 1 mol/ 35.5 g = 18.3 mole Cl

mol C2H6/
mol Cl= 1/2

mol C2H6/ mol Cl = 10 g/18.3 g = .55 (Cl is the limiting reagent)


18.3 mol Cl X 1 mol C2H5Cl/ 2 mol Cl =
9.15 mol C2H5Cl


9.15 mol C2H5Cl X 64.5 g C2H5Cl/ 1 mol c2H5Cl =
590g C2H5Cl (Theoretical Yeild)


Actual/Theoretical = 490g / 590g =
83% yeild

This is so boring.

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