Question

What volume will 454 g (1 lb.) of H₂ occupy at 1.05 atm and 25.0°c

Answer 1

V
=
5.24
×
10
3

L
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

Answer 2

Answer:

Explanation:

V = 5.24 × 10^3L