It looks like you may not have converted the mass of #"H"_2"# to moles #(n)#. Also, you have the incorrect value for #R# because you are using mmHg as the pressure unit. The value you were using is what you would use if the pressure was in atm. You could convert from mmHg to atm.
The equation for the ideal gas law is:
#PV=nRT#,
where #P# is pressure, #V# is volume, #n# is moles, #R# is a gas constant, and #T# is temperature in Kelvins.
You have to convert the mass into moles. You do that by multiplying the given mass of #"H"_2"# by its molar mass which is #"2.016 g/mol"#, which is 2 times the molar mass of #"H"_2"#, which is its atomic weight in grams/mole, or g/mol.
#2.50color(red)cancel(color(black)("g H"_2))xx(1"mol H"_2)/(2.016color(red)cancel(color(black)("g H"_2)))="1.24 mol H"#
#color(blue)("Organize you data"#.
Given/Known
#P="828 mmHg"# #lArr# You also gave a pressure of 820 mmHg. If its supposed to be 820, you can recalculate using that value.
#n="1.24 mol"#
#R="62.36367 L mmHg K"^(-1) "mol"^(-1)"#
http://www.cpp.edu/~lllee/gasconstant.pdf
#T="308 K"#
Unknown: #V#
#color(blue)("Solution"#
Rearrange the equation to isolate #V#. Insert your data and solve.
#V=(nRT)/P#
#V=(1.24color(red)cancel(color(black)("mol"))xx62.36367color(white)(.) "L" color(red)cancel(color(black)("mmHg")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx308color(red)cancel(color(black)("K")))/(828color(red)cancel(color(black)("mmHg")))"="28.8color(white)(.)"L H"_2"#
Just in case you meant #"820 mmHg"#, you would get the following:
#V=(1.24color(red)cancel(color(black)("mol"))xx62.36367color(white)(.) "L" color(red)cancel(color(black)("mmHg")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx308color(red)cancel(color(black)("K")))/(820color(red)cancel(color(black)("mmHg")))"="29.0color(white)(.)"L H"_2"#
