Ok, I think I might have just had a lightbulb moment.
(138 g H2SO4)/(98 g H2SO4/mol)/(8 moles S8)x(256 g S8)= 45.1 g S8
=45.1g S8 <---can someone confirm????
Elemental sulfur can form sulfur trioxide when burned. When sulfur trioxide is dissolved in water, it yields sulfuric acid. The net reaction is shown. What mass of sulfur is needed to yield 138 grams of sulfuric acid.
S8(s) + 12O2(g) + 8H2O(l) --> 8H2SO4(l)
Help?
3 answers
The number is correct but look at the algebra. I don't think you have 8 mols S8; you have 1 mol S8. Check this out.
mols H2SO4 = 138/98. OK.
Convert to mols S8. That will be
138/98 x (1 mol S8/8 mol H2SO4).
Now convert that to g S8
138/98 x (1/8) x (molar mass S8) =
138/98 x (1/8) x (8*32) = 45.1
If I went through your algebra correctly, you made an error in algebra which just happened to produce the correct answer (with the 8S8 in your answer).
mols H2SO4 = 138/98. OK.
Convert to mols S8. That will be
138/98 x (1 mol S8/8 mol H2SO4).
Now convert that to g S8
138/98 x (1/8) x (molar mass S8) =
138/98 x (1/8) x (8*32) = 45.1
If I went through your algebra correctly, you made an error in algebra which just happened to produce the correct answer (with the 8S8 in your answer).
(138 g H2SO4)/(98 g H2SO4/mol)/(8 moles S8)x(256 g S8)= 45.1 g S8 Should have been:
(138 g H2SO4)/(98 g H2SO4/mol)/(8 moles {{{H2SO4}}})x(256 g S8)= 45.1 g S8
I typed S8 by mistake instead of H2SO4, I had H2SO4 on my paper.
Thanks for your help.
(138 g H2SO4)/(98 g H2SO4/mol)/(8 moles {{{H2SO4}}})x(256 g S8)= 45.1 g S8
I typed S8 by mistake instead of H2SO4, I had H2SO4 on my paper.
Thanks for your help.