When a mixture of 10 cm^3 of oxygen and 50 cm^3 of hydrogen is sparked continuously, what is the maximum theoretical decrease in volume?
Answers
This one is a little tricky. And here's why I say that.
When it comes to reaction that involve gases kept under the same conditions for pressure and temperature, it's important to realize that the mole ratios that exist between the species involved in the reaction becomes equivalent to the volume ratio.
So, take a look at the balanced chemical equation for your reaction
Notice that you have a
2
:
1
mole ratio between carbon monoxide and oxygen gas. This tells you that the reaction will always consume twice as many moles o carbon monoxide than of oxygen gas.
But since we're working with gases under the same conditions for pressure an temperature, you can say the exact same thing about their volumes.
In other words, the reaction will always consume a volume of carbon monoxide that is twice as large as the volume of oxygen gas.
Now, here's where the tricky part comes in. Notice that the problem provides you with equal volumes of
CO
and
O
2
.
This is a problem because you know that you'd need a volume of
CO
that is twice as large as the volume of
O
2
.
You an thus conclude that you're dealing with a limiting reagent. More specifically, carbon monoxide will act as a limiting reagent because it will be consumed before all the oxygen gets a chance to react.
So, you can say that the reaction will only consume
10
cm
3
CO
⋅
1 cm
3
O
2
2
cm
3
CO
=
5 cm
3
a
O
2
So,
10 cm
3
of
CO
wil lreact with
5 cm
3
of
O
2
, leaving
5 cm
3
of
O
2
in excess.
Now look at the
2
:
2
volume ratio that exists between
CO
and
CO
2
. This tells you that for a given volume of
CO
that takes part in the reaction, the reaction produces an equal volume of
CO
2
.
So, if
10 cm
3
of
CO
react, you will get
10 cm
3
of
CO
2
produced.
This means that after the reaction is finished, the reaction vessel will contain
V
gas
=
in excess
V
O
2
+
produced
V
C
O
2
V
gas
=
5 cm
3
+
1 0cm
3
=
15 cm
3