metalayer - 28-11-2014 at 08:12
Can someone check I have this right?
I need a 1.5:1 molar ratio of 6% NaOCl to 20g of an alkene with a MW of 207.25g/mole.
- Make 1l 6% NaOCl - 428.5ml of 14% stock in 571.5ml h2o.
- Need to calculate moles of solute in stock 6% NaOCl first. 6% of 1000 = 60g NaOCl = 0.8 moles.
- Then calculate molar ratio based on 1l = 1.5:1 = 0.8:0.5 = 1l 6% NaOCl:103.6g alkene.
- Scale down for 20g = 193ml:20g
blogfast25 - 28-11-2014 at 09:41
20 g of the alkene is 20/207.25 = 0.0965 mol alkene.
Molar ratio NaOCl to alkene is 1.5:1, so you need 0.0965 x 1.5 = 0.145 mol pure NaOCl or 0.145 mol x 74.442 g/mol = 10.8 g pure NaOCl.
6 g pure NaOCl is contained in 100 g of NaOCl solution, so 10.8 g is contained in 100 x 10.8/6 = 180 g of 6 % NaOCl solution.
[Edited on 28-11-2014 by blogfast25]
metalayer - 28-11-2014 at 10:34
Thanks, I see where I went wrong now.