Gas Stoichoimetry dealing with Mole Ratio

[pepper.mix.salt]

hey please help!

this question really got me stumped.

MxOy a gaseous ocide of an element M gives an acidic gas, MO2 on combustion with excess O2.

When the combustion is carried out between 10cm3 of MxOy and 40cm3 of O2 (in excess), the total volume of gas (after cooling to the original temperature) is unchanged but on shaking with some aqeous KOH, the volume is reduced to 20cm3. All volumes are measured at the same temperature and pressure.

what are the values of x and y?

please help, thank you! :(

[ILoveMaths07]

MxOy + O2 ---> MO2 + H2O

This is the unbalanced equation for the reaction.

10 cm^3 of MxOy react with 40 cm^3 of O2 in excess. All the MxOy will react, but all the O2 won't. How do you know how much O2 reacts?
After shaking with some KOH, the volume of gas is reduced to 20 cm^3. It was initially 40 cm^3 (40 cm^3 of excess oxygen). KOH, a base, will only react with MO2, the acidic gas. If the volume of gas is reduced to 20 cm^3 AND the total volume of gas is still 40 cm^3, it means that 20 cm^3 of MO2 is formed, and 20 cm^3 of O2 reacted.

Now look at the ratio of the volumes of each reagent.

MxOy = 10 cm^3
O2 = 20 cm^3 (We don't take 40 cm^3, because we only need the volume of oxygen that reacted)
MO2 = 20 cm^3

The ratios are 1:2:2. Let these be the stoichiometric coefficients in the equation.

MxOy + 2O2 ---> 2MO2 + H2O

We have four oxygens on the left hand side, and five oxygens on the right. The left hand side needs one more oxygen, so y = 1.
We have two M's on the right hand side, so x = 2.

The balanced equation is M2O + 2O2 ---> 2MO2 + H2O.

I hope that helps. :)

ILoveMaths07.

P.S. Sorry about that. I don't think there's any water formed in this reaction. If I don't include water in the equation, then y = 0, which is ridiculous. Are you sure you've copied down the volumes correctly? Is the question fine? But that's how you work out the problem.