We need to determine the moles "O"_2" under the conditions given. Then we need to determine the mass of "O"_2" of the calculated moles. Then we will subtract the calculated mass of "O"_2" from "305 g O"_2" to determine the mass of "O"_2" that must be released.
First use the equation for the ideal gas law to determines moles "O"_2" at 1.15"atm":
PV=nRT,
where:
P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvins.
Organize the data:
Known/Given
P="1.15 atm"
V="34.0 L"
R="0.0820575 L atm K"^(-1) "mol"^(-1)"
T="22"^@"C + 273.15 = 295 K"
Unknown
n
Determine moles "O"_2".
Rearrange the equation to isolate n. Plug in the known values and solve.
n=(PV)/(RT)**
n=(1.15color(red)cancel(color(black)("atm"))xx34.0color(red)cancel(color(black)("L")))/(0.0820575 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) color(red)cancel(color(black)("K"))^(-1) "mol"^(-1)xx295color(red)cancel(color(black)("K")))="1.6152 mol O"_2"
(I am keeping some guard digits to reduce rounding errors.)
Convert moles "O"_2" to mass "O"_2".
Multiply moles "O"_2" by its molar mass ("31.998 g/mol").
1.6152color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="51.7 g O"_2" (rounded to three significant figures)
The amount of "O"_2" that must be released:
"305 g O"_2-"51.7 g O"_2"= 253 g O"_2" (rounded to a whole number)
Under the conditions given, "253 g O"_2" need to be released to reduce the pressure to "1.15 atm".