20ml of carbon dioxide and ethane was exploded with 50ml of oxygen gas. the volume after explosion was 45m, shaking with sodium hydroxide solution only 15ml of oxygen gas was left behind. what were the volume of carbon monoxide and ethane in mixture
Answers
Start by writing a balanced chemical equation for your reaction
CH
4
(
g
)
+
2
O
2
(
g
)
→
CO
2
(
g
)
+
2
H
2
O
(
l
)
Notice that I added water as a liquid,
(
l
)
, because the problem tells you that after the reaction is complete, the resulting gaseous mixture is cooled down to room temperature.
Now, notice that the reaction consumes
2
moles of oxygen gas for every
1
mole of methane that takes part in the reaction and produces
1
mole of carbon dioxide.
When your reaction involves gases kept under the same conditions for pressure and temperature, you can treat the mole ratios that exist between them in the balanced chemical reaction as volume ratios.
In your case, you can say that the reaction consumes
2 mL
of oxygen gas for every
1 mL
of methane that takes part in the reaction and produces
1 mL
of carbon dioxide.
This means that in order for the reaction to consume all the methane present in the sample, you need
20
mL CH
4
⋅
2 mL O
2
1
mL CH
4
=
40 mL O
2
As you can see, you have more oxygen gas than you need to ensure that all the methane reacts
→
oxygen gas is in excess, which is equivalent to saying that methane is a limiting reagent.
So, the reaction will consume
20 mL
of methane and
40 mL
of oxygen gas and produce
20
mL CH
4
⋅
1 mL CO
2
1
mL CH
4
=
20 mL CO
2
After the reaction is complete, you will be left with
50 mL O
2
−
40 mL O
2
=
10 mL O
2
that are not consumed by the reaction and with
20 mL
of carbon dioxide. Therefore, you can say that after the reaction is complete, your mixture will contain
10 mL O
2
+
20 mL CO
2
=
30 mL gas