This can proven using the ideal gas law equation for two gases kept at a temperature
T
and a pressure
P
.
P
⋅
V
1
=
n
1
⋅
R
⋅
T
→
the ideal gas law equation for the first gas
P
⋅
V
2
=
n
2
⋅
R
⋅
T
→
the ideal gas law equation for the second gas
Divided these two equations to get
P
⋅
V
1
P
⋅
V
2
=
n
1
⋅
R
⋅
T
n
2
⋅
R
⋅
T
This means that you have
n
1
n
2
=
V
1
V
2
→
the mole ratio is equivalent to the volume ratio
In your case, the balanced chemical equation looks like this
N
2(g]
+
3
H
2(g]
→
2
NH
3(g]
The problem tells you that this reaction takes place at constant temperature and pressure. You don't need to know the exact values, you just need to know that they are constant for all three chemical species.
You can thus say that since you have a
1
3
mole ratio between nitrogen gas and oxygen gas, you will also have a
1
3
volume ratio between these two gases
n
N
2
n
H
2
=
V
N
2
V
H
2
=
1
3
The volume of hydrogen gas needed to react with that much nitrogen gas will thus be
15l is answer