Asked by Scott
If 25.0 g S8 (molar mass = 256.56 g/mol) is made to react with 2.0 L O2 at 25 oC and
1.0 atm, what is the maximum amount (in grams) of SO3 produced?
1.0 atm, what is the maximum amount (in grams) of SO3 produced?
Answers
Answered by
scott
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
moles of O ... 2 * 2.0 / 22.4
O will be the limiting reactant
be aware of significant figures
Answered by
Scott
Just want to double check, I ended up getting 4.363g SO3
Answered by
scott
your number seems low
4.76 g ?
4.76 g ?
Answered by
Scott
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
Then i did (0.081748 mol O2) x (8 mol SO3 / 12 mol O2) x (80.0632 g SO3/mol) = 4.36 g SO3
Answered by
scott
okay ... I didn't make the stp correction for the O2
Answered by
Scott
Np thanks scott for the help
Answered by
I'm going to cry
why are you both named scott?
-rhetorical question
-rhetorical question
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