This can be solved by using the Ideal Gas Law:
PV=nRTPV=nRT,
where PP is pressure, VV is volume, nn is moles, RR is the universal gas constant, and TT, which is the temperature in Kelvins.
First calculate the number of moles of "O"_2"O2, using the equation for the Ideal Gas Law. Then multiply the moles "O"_2O2 by its molar mass to get the mass of "O"_2"O2.
color(blue)("Organize your data"Organize your data
Known/Given
P="8.16 atm"P=8.16 atm
V="5.60 L"V=5.60 L
R="0.08206 L atm K"^(-1) "mol"^(-1)R=0.08206 L atm K−1mol−1
T="23"^@"C"+273="296 K"T=23∘C+273=296 K
color(blue)("Moles of Oxygen Gas"Moles of Oxygen Gas
Rearrange the equation to isolate nn. Insert your data and solve.
n=(PV)/(RT)n=PVRT
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