Thanks UG I’ve done it a couple of ways now and I got the same answer, thanks for going to all the trouble to reply Fled. I didn’t realise you wanted me to identify all the units to do the factor label method. Where I’m from in Australia that method is not really taught at the High School or technician level, though it probably is at university level I’m not sure.
Actually I’m not a high school student I just wanted someone to verify an answer from a peculiar situation that came up at work so I may have put this in the wrong section.
How about this:
Al = 1.4mg/L
Al = 0.0014g/L
n moles (Al): 0.0014g/L / 26.98g/Mole = 0.00005189 M/L
2 moles Al : 1 mole Al2(SO4)3
No moles Al2(SO4)3 = 0.00005189/2
= 0.000025945 M/L
Conc Al2 (SO4)3 = 0.000025945 x 342.14g
I think my issue is similar to this problem: http://au.answers.yahoo.com/question/index?qid=20110327231539AAJJ5RE
I’m getting an answer of 1.026g/50ml for that one so I think I'm now on the right track.
Just to recap the question
Al = 1.4 mg/L
What is the conc of Al2(SO4)3
If anyone wants to give it a go to verify.