Asked by Mike
The question is: Find the following quantities
the volume, in liters, of 835 g SO3 at STP
My answer is:
1 mol of SO3=80g
835g/80g = 10.44 mols
v = (10.44 x 8.314 x 273)/100 = 236 L
Can someone tell me if I did it correct. Thanks very much.
the volume, in liters, of 835 g SO3 at STP
My answer is:
1 mol of SO3=80g
835g/80g = 10.44 mols
v = (10.44 x 8.314 x 273)/100 = 236 L
Can someone tell me if I did it correct. Thanks very much.
Answers
Answered by
DrBob222
I think you should have used 101.3 kPa for pressure and not 100, then the answer would have been about 234 L. An easier way is to remember a mole of a gas at STP occupies 22.4 L; therefore,
10.44 mol x (22.4 L/mol) = about 234.. I would also suggest that if you have 3 significant figures in the 835 that you should have used at least 3 places in the molar mass of SO3. I would have used 80.064, don't clear the calculator, multiply by 22.4, then round to 3 places.
10.44 mol x (22.4 L/mol) = about 234.. I would also suggest that if you have 3 significant figures in the 835 that you should have used at least 3 places in the molar mass of SO3. I would have used 80.064, don't clear the calculator, multiply by 22.4, then round to 3 places.
Answered by
Mike
Thank you. So I would have:
1 mol of So3 = 80.064
835g / 80.064g = 10.44
v = (10.44 x 80.064 x 273) / 101.3 = 234 L
1 mol of So3 = 80.064
835g / 80.064g = 10.44
v = (10.44 x 80.064 x 273) / 101.3 = 234 L
Answered by
DrBob222
If you wished to use the ideal gas law, yes.
Answered by
Mike
Thank you so much for your time and explanation
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