Question

Under conditions of `"NTP"`, what volume of `SO_2(g)` results from burning a `10*g` mass of elemental sulfur?

Answer 1

We address the stoichiometric equation...........and I get a volume of under `10*L`...............

Explanation:

`S(s)+O_2(g) rarrSO_2(g)`

Now depending on your definition of `"NTP"` (they seem to change with the wind across syllabuses) we can work out the volume. I use values of `"NTP"` to specify `P=1*atm`, and `T=293.15*K`.

`"Moles of sulfur "-=" Moles of sulfur dioxide"`

`"Moles of sulfur"-=(10*g)/(32.06*g*mol^-1)=0.312*mol`.

And thus there are `0.312*mol` `SO_2` gas........given conversion stoichiometric conversion.

`V=(nRT)/P=(0.312*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(1*atm)`

`=7.5*L`