Answer:
[tex]4.18LH_2[/tex]
Explanation:
The important thing to keep in mind when dealing with gases that are under the same conditions for pressure and temperature is that their mole ratio is equivalent to their volume ratio.
You can prove this by using the ideal gas law equation
P
⋅
V
1
=
n
1
⋅
R
T⇒ the first gas at pressure P and temperature T
P
⋅
V
2
=
n
2
⋅
R
T
⇒ the second gas at pressure P and temperature T
Divide these two equations to get
[tex]\frac{PV_1}{PV_2} =\frac{n_1RT}{n_2RT}[/tex]
Therefore, you can say that
[tex]\frac{n_1}{n_2} = \frac{V_1}{V_2}[/tex] the mole ratio is equal to the volume ratio
The balanced chemical equation for your reaction looks like this
2
H
2(g]
+
O
2(g]
→
2
H
2
O
(g]
4.55
L O
2
⋅
2
L H
2
1
L O
2
=
9.10 L H
2
[tex]= 2.09LO_2 .\frac{2LH_2}{1LO_2} \\=4.18LH_2[/tex]
