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Chemistry gas pressure alevel

Could someone help me as I keep getting this question wrong 😐
1 . Warning a gaseous oxide of chlorine ClxOy causes it to decompose
2clxOy(g) + xCl2(g) + yO2(g)
In the experiment 40cm3 of the oxide gave 40cm3 of the chlorine and 20cm3 of the oxygen. This is all at the same pressure and temp. Could someone just help with working out the Y as I’ve worked out X as 2
Balanced equation = 2ClxOy = 2Cl2 + O2
Original post by Matildamcafee
Could someone help me as I keep getting this question wrong 😐
1 . Warning a gaseous oxide of chlorine ClxOy causes it to decompose
2clxOy(g) + xCl2(g) + yO2(g)
In the experiment 40cm3 of the oxide gave 40cm3 of the chlorine and 20cm3 of the oxygen. This is all at the same pressure and temp. Could someone just help with working out the Y as I’ve worked out X as 2
Balanced equation = 2ClxOy = 2Cl2 + O2

You appear to have a problem with typing as well.

I don't know any compounds that decompose when you warn them. Rather too sensitive to authority, I think.

Let me have a go at writing the question:

1 . Warming a gaseous oxide of chlorine ClxOy causes it to decompose.

2ClxOy(g) ==> xCl2(g) + yO2(g)

In the experiment 40cm3 of the oxide gave 40cm3 of the chlorine and 20cm3 of the oxygen. This is all at the same pressure and temp. Could someone just help with working out the Y as I’ve worked out X as 2


Avogadro's law says that the volume is proportional to the number of moles in all gases (at constant T and P).
So, if 40cm3 of the oxide represents 2ClxOy
and 40cm3 of chlorine is produced,
then mol of chlorine = mol of oxide
hence x = 2

If only 20cm3 of oxygen are produced (half the volume of the initial oxide)
then mol oxygen = mol oxide/2
hence, y = 1

Cl2O is the unknown oxide

Balanced equation: 2Cl2O ==> 2Cl2 + O2
Original post by charco
You appear to have a problem with typing as well.

I don't know any compounds that decompose when you warn them. Rather too sensitive to authority, I think.

Let me have a go at writing the question:

1 . Warming a gaseous oxide of chlorine ClxOy causes it to decompose.

2ClxOy(g) ==> xCl2(g) + yO2(g)

In the experiment 40cm3 of the oxide gave 40cm3 of the chlorine and 20cm3 of the oxygen. This is all at the same pressure and temp. Could someone just help with working out the Y as I’ve worked out X as 2


Avogadro's law says that the volume is proportional to the number of moles in all gases (at constant T and P).
So, if 40cm3 of the oxide represents 2ClxOy
and 40cm3 of chlorine is produced,
then mol of chlorine = mol of oxide
hence x = 2

If only 20cm3 of oxygen are produced (half the volume of the initial oxide)
then mol oxygen = mol oxide/2
hence, y = 1

Cl2O is the unknown oxide

Balanced equation: 2Cl2O ==> 2Cl2 + O2

I asked for help with the question, not unnecessary rudeness. I’m 16, I’m sure an adult can look past a typo.
Original post by Matildamcafee
I asked for help with the question, not unnecessary rudeness. I’m 16, I’m sure an adult can look past a typo.

de nada ...

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