[elitewarr]

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.

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?

thanks.

[Borek]

Why do you think water was removed by KOH solution?

What is T of RTP?

[elitewarr]

I think that water is removed since it will just dilute the KOH and remain in the solution.

From the question, the temperature is at room temperature. But if that is so, then water is formed. But if water in liquid form is formed, then everything is even more complicated. Even so, if i use the density 1.0g/cm^3 of water to calculate, it just does not make sense.