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
