ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: luffy on March 14, 2011, 04:47:31 pm
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Hi, guys:
When I did the question below, my answer was different to the one given in the textbook. I just wanted to know whether or not I was completing it correctly.
A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500g sample of aluminium sulfate (Al2(SO4)3) were dissolved in a volume of water and excess barium chloride added to precipitate barium sulfate. What was the total mass of barium sulfate produced?
Thanks in advance.
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since you only want the answer.
I'm too lazy so here's my answer
m(BaSO4)=1.8445g
I'll post the method if you want them, just ask
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since you only want the answer.
I'm too lazy so here's my answer
m(BaSO4)=1.8445g
I'll post the method if you want them, just ask
Could you please show me how to do it too? I must have made an error somewhere.
Thanks.
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I got 1.8445
 = \frac{0.5}{142.06} = 0.00352 mol )
 in sample = 0.00352 x 96.06 = 0.338097g)
_3)) = 0.5/342.14 = 0.001461 mol)
 in sample = 0.004384 mol)
 in sample = 0.004384 x 96 = 0.421143g)
n(Ba) which reacted with Soldium sulfate = 0.00353 mol
therfore m(Ba) = 0.00352 x 137.3 = 0.455836g
n(Ba) which reacted with Aluminium sulfate = 0.004384 mol
therefore m(Ba) = 0.004384 x 137. 3 = 0.60192 g
add them altogether to find the mass of BaSO4 and you get 1.8445
Edit: ceebs applying latex, also, how do i insert spaces in my latex equations?
Double edit: Btw, i used an overly-complicated long way to do this, you should just work out the total moles of
then with that find the total mols of Barium, then find the mass of the precipitate. Sorry
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lol just google it.
But I think that's it. havent tested it yet though.
http://www.personal.ceu.hu/tex/spacebox.htm