50cm^3 of a mixture of CO, CO2 and H2 were exploded with 25.0cm^3

[elitewarr]

Homework Statement

50cm^3 of a mixture of CO, CO2 and H2 were exploded with 25.0cm^3 of O2. After explosion, the volume measured at r.t.p. was 37.0 cm^3. After treatment with aqueous KOH, the volume was reduced to 5.0cm^3. Calculate the % composition by volume of the original mixture.

Homework Equations

The Attempt at a Solution

From this, I can safely assume that CO will react with O2 to form CO2
2CO + O2 --> 2CO2
and
2H2 + O2 --> 2H2O
and CO2 will not react with Oxygen.
When treated with KOH, the steam and carbon dioxide will all be removed. I assumed that 5.0cm^3 of oxygen is left. However, what I don't understand is how did the total volume of gas decrease from 75cm^3 (including oxygen) to 37.0cm^3?

 

[chemisttree]

Is there any steam at r. t. p.? Water vapor? How much water vapor will there be after treatment with KOH? More? Less? The same? I assume that the KOH treatment is an aqueous solution of KOH but your instructor might have a different idea.

...However, what I don't understand is how did the total volume of gas decrease from 75cm^3 (including oxygen) to 37.0cm^3?

Three gases at the beginning. How many gases at the end (at r. t. p.)?

 

[elitewarr]

Hmm. Good question. Is there steam? If there isn't, then everything becomes complicated.
I guess the 'steam' will just stay with the KOH after treatment so that oxygen is the only gas that is left.
So, at the end, there is oxygen.
But I'm not that sure about that. Maybe not all carbon monoxide or hydrogen is reacted, so they may be collected? But, if I think it like that, I'm just complicating things.