# Volume of gases question

If 100cm³ Hydrogen reacts with 700cm³ Oxygen (an excess), what will be the
total volume of gases after the reaction?

H2(g)+202(g)O3(g) + H2O(l)
Original post by Lucy_Alice_1
If 100cm³ Hydrogen reacts with 700cm³ Oxygen (an excess), what will be the
total volume of gases after the reaction?

H2(g)+202(g)O3(g) + H2O(l)

Firstly, what have you tried so far?

Secondly, that equation won’t help as it is not balanced nor should it show both ozone and oxygen on the left hand side. (Edit: I now realise that aside from the lack of an arrow between the 2O2 (g) and O3 (g), the equation is perfectly correct, it just represents a reaction that isn’t especially likely to occur)
(edited 9 months ago)
Original post by Lucy_Alice_1
If 100cm³ Hydrogen reacts with 700cm³ Oxygen (an excess), what will be the
total volume of gases after the reaction?

H2(g)+202(g)O3(g) + H2O(l)

Can't you just the volumes as these are equivalent to moles at the same T and P. So all hydrogen used and use twice as much oxygen I.e 200cm3. So 500cm3 left. 100 cm3 of ozone formed as 1:1wuth hydrogen. Total vol is 500 +100=600cm3
Original post by Mole man
Can't you just the volumes as these are equivalent to moles at the same T and P. So all hydrogen used and use twice as much oxygen I.e 200cm3. So 500cm3 left. 100 cm3 of ozone formed as 1:1wuth hydrogen. Total vol is 500 +100=600cm3

Firstly, as pointed out above, you need to identify the correct equation for combustion of hydrogen (hint; there is no ozone involved).
Original post by charco
Firstly, as pointed out above, you need to identify the correct equation for combustion of hydrogen (hint; there is no ozone involved).

I agree! I just presumed that the equation had been given in the question!