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"#.
