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????
S8(s) + 12O2(g) + 8H2O(l) --> 8H2SO4(l)
Help?
(138 g H2SO4)/(98 g H2SO4/mol)/(8 moles S8)x(256 g S8)= 45.1 g S8
=45.1g S8 <---can someone confirm????
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 {{{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.
First, let's compare the balanced equation to the stoichiometry of sulfur. From the equation, we see that 1 mole of S8 reacts to produce 8 moles of H2SO4.
Next, we need to find the molar mass of H2SO4. The molar mass of H2SO4 is calculated as follows:
(2 x molar mass of hydrogen) + molar mass of sulfur + (4 x molar mass of oxygen)
= (2 x 1.01 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
= 98.09 g/mol
Now, we can set up a proportion to find the amount of sulfur needed:
(1 mole of S8 / 8 moles of H2SO4) = (mass of sulfur / 138 g)
Cross-multiplying:
mass of sulfur = (1 mole of S8 / 8 moles of H2SO4) * 138 g
Finally, convert the moles of S8 to grams:
molar mass of S8 = 256.54 g/mol (8 x 32.07 g/mol)
mass of sulfur = [(1 mole of S8 / 8 moles of H2SO4) * 138 g] * (molar mass of S8 / 1 mole of S8)
= (138 g * 256.54 g/mol) / (8 * 98.09 g/mol)
Calculating the final answer:
mass of sulfur = 569.7 g (rounded to one decimal place)
Therefore, approximately 569.7 grams of sulfur are needed to yield 138 grams of sulfuric acid.