We need to determine the moles "O"_2"O2 under the conditions given. Then we need to determine the mass of "O"_2"O2 of the calculated moles. Then we will subtract the calculated mass of "O"_2"O2 from "305 g O"_2"305 g O2 to determine the mass of "O"_2"O2 that must be released.
First use the equation for the ideal gas law to determines moles "O"_2"O2 at 1.15"atm"1.15atm:
PV=nRT,PV=nRT,
where:
PP is pressure, VV is volume, nn is moles, RR is the gas constant, and TT is temperature in Kelvins.
Organize the data:
Known/Given
P="1.15 atm"P=1.15 atm
V="34.0 L"V=34.0 L
R="0.0820575 L atm K"^(-1) "mol"^(-1)"R=0.0820575 L atm K−1mol−1
T="22"^@"C + 273.15 = 295 K"T=22∘C + 273.15 = 295 K
Unknown
nn
Determine moles "O"_2"O2.
Rearrange the equation to isolate nn. Plug in the known values and solve.
n=(PV)/(RT)n=PVRT**
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".