Question

What mass of HCl (in Kg) is required to create 8.80 Liters of hydrogen gas?

Answer 1

The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
.
Moreover, at STP one mole of any ideal gas occupies exactly
22.7 L
- this is known as the molar volume of a gas at STP.
Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.
To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as
P
V
=
n
R
T
⇒
n
=
P
V
R
T
For hydrogen, you would have
n
hydrogen
=
P
⋅
V
hydrogen
R
T
and for chlorine you have
n
chlorine
=
P
⋅
V
chlorine
R
T
Thus, the mole ratio between hydrogen and chlorine will be
n
hydrogen
n
chlorine
=
P
V
hydronge
R
T
⋅
R
T
P
⋅
V
chlorine
=
V
hydrogen
V
chlorine
The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.
So, the balanced chemical equation for this reaction is
H
2(g]
+
Cl
2(g]
→
2
HCl
(g]
Notice that you have a
1
:
2
mole ratio between hydrogen gas and hydrogen chloride.
This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.
Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction
4.9
L H
2
⋅
2
L HCl
1
L H
2
=
9.8 L HCl
Now use the molar volume to find how many moles you'd get in this volume of gas at STP
9.8
L HCl
⋅
1 mole HCl
22.7
L HCl
=
0.4317 moles HCl
Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles
0.4317
moles HCl
⋅
36.461 g
1
mole HCl
=
15.74 g
Rounded to two sig figs, the answer will be
m
HCl
=
16 g