If 25.0 g S8 (molar mass = 256.56 g/mol) is made to react with 2.0 L O2 at 25 oC and

Question

1.0 atm, what is the maximum amount (in grams) of SO3 produced?

Answer 1

moles of S ... 8 * 25 / 256.56

moles of O ... 2 * 2.0 / 22.4

O will be the limiting reactant

be aware of significant figures

Answer 2

Just want to double check, I ended up getting 4.363g SO3

Answer 3

your number seems low

4.76 g ?

Answer 4

First i did (1 atm x 2 L) / (K(constant) x 298.15K) = 0.081745mol

Then i did (0.081748 mol O2) x (8 mol SO3 / 12 mol O2) x (80.0632 g SO3/mol) = 4.36 g SO3

Answer 5

okay ... I didn't make the stp correction for the O2

Answer 6

Np thanks scott for the help

Answer 7

why are you both named scott?
-rhetorical question