V
=
5.24
×
10
3
L
Step-by-step explanation:
We're asked to find the volume occupied by
454
g
of
H
2
at
1.05
atm
and
25
o
C
.
To do this, we can use the ideal-gas equation:
p
V
=
n
R
T
where
p
is the pressure, in units of
atm
(given as
1.05
atm
)
V
is the volume the gas occupies, in units of
L
(we'll be finding this)
n
is the number of moles of the gas present (we'll need to convert the given mass in grams to moles)
R
is the universal gas constant, equal to
0.082057
L
⋅
atm
mol
⋅
K
T
is the absolute temperature of the gas, in units of
K
(given
25
o
C
)
We need to do some conversions:
Let's find the number of moles of
H
2
present via the molar mass of
H
2
(
2.02
g/mol
):
454
g H
2
(
1
l
mol H
2
2.02
g H
2
)
=
225
mol H
2
The temperature in
K
is
25
o
C
+
273
=
298
K
Plugging in known values, and solving for the volume,
V
, we have
V
=
n
R
T
p
=
(
225
mol
)
(
0.082057
L
⋅
atm
mol
⋅
K
)
(
298
K
)
(
1.05
atm
)
=
5.24
×
10
3