What is the mass of #"5.6 L"# of #"O"_2"# at #23^@"C"# and #"8.16 atm"#?

1 Answer
Jun 1, 2017

Answer:

The mass of #"O"_2# is #"60.2 g"#

Explanation:

This can be solved by using the Ideal Gas Law:

#PV=nRT#,

where #P# is pressure, #V# is volume, #n# is moles, #R# is the universal gas constant, and #T#, which is the temperature in Kelvins.

First calculate the number of moles of #"O"_2"#, using the equation for the Ideal Gas Law. Then multiply the moles #"O"_2# by its molar mass to get the mass of #"O"_2"#.

#color(blue)("Organize your data"#

Known/Given

#P="8.16 atm"#

#V="5.60 L"#

#R="0.08206 L atm K"^(-1) "mol"^(-1)#

#T="23"^@"C"+273="296 K"#

#color(blue)("Moles of Oxygen Gas"#

Rearrange the equation to isolate #n#. Insert your data and solve.

#n=(PV)/(RT)#

#n=(8.16color(red)cancel(color(black)("atm"))xx5.60color(red)cancel(color(black)("L")))/(0.08206 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx296color(red)cancel(color(black)("K")))="1.88 mol"#

There are #"1.88 mol O"_2"# under the conditions described in the question.

#color(blue)("Mass of Oxygen Gas"#

Multiply the mol #"O"_2"# by its molar mass, #"31.998 g/mol"#.

#1.88color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="60.2 g O"_2# rounded to three sig figs