The volume of 9.00 grams of carbon dioxide vapor at STP is 11.2 L. How would I figure that out?

1 Answer
May 23, 2018

Answer:

Just as an exercise...., and I realize this is an OLD question, first you should specify #"STP"#..

Explanation:

All sorts of standards exist across different curricula that specifies standard conditions. At the moment it really is a dog's breakfast, and the situation is much more confused than when I was a student.

Following this site, which may disagree with your curriculum, #"STP"=273.15*K, 1*"bar"#...#"1 bar"-=0.987*atm#...I myself like using atmospheres, as it is an intuitive unit of pressure, and I suspect that most chemists of my and earlier generations would agree.

And so we solve the Ideal Gas equation....#PV=nRT#

#V=(nRT)/P=((9.00*g)/(44.01*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx273.15*K)/(0.987*atm)#

And I make this...#V=4.65*L#...so I think there are unspecified data in your question....or I (or the question) might have totally ballsed it up.