Question

Q.)(i) How many moles of sulphur will be produced when 2 moles of H2S reacts with 11.2L of SO2at NTP?

Answer 1

I think you refer to the so-called `"comproportionation reaction"`....

Explanation:

The sulfide in `"hydrogen sulfide"` is OXIDIZED `(i)` to elemental sulfur, and the `"sulfur dioxide"` is REDUCED to elemental sulfur......`(ii)`

`S^(2-) rarrS +2e^(-)` `(i)`

`SO_2 +4H^(+) + 4e^(-) rarrS+2H_2O` `(ii)`

Both mass and charge are balanced in each reaction, as indeed they must be if we purport to represent chemical reality.....and so we simply add `2xx(i) + (ii)` to eliminate the electrons to give........

`2S^(2-) + SO_2 +4H^(+) + cancel(4e^(-)) rarr3S+2H_2O+cancel(4e^(-))`

And we could simplify further to give.......

`2H_2S + SO_2 rarr3S+2H_2O`

And now (finally) we can address your question. `"NTP"` specifies a molar volume of `24.06*L*mol^-1`......

With respect to `SO_2` we have a molar quantity of `(11.2*L)/(24.06*L*mol^-1)=0.466*mol....`. And thus `SO_2` is the limiting reagent and `"hydrogen sulfide"` is in excess.

And so we gets `3 xx` an equivalent quantity of `S`, i.e. `(3 * S)/(1 * SO_2) xx 0.466*molsxx32.06*g*mol^-1=44.8*g`.

And even if you did this reaction carefully in a well-ventilated hood, it would still stink, and the upstairs chemists would come down to give you a beatdown......

Answer 2

0.188 mole `S_8(s)`

Explanation:

`16H_2S(g) + 8SO_2(g)` => `16H_2O(l)` + `3S_8(s)`
Since the form of sulfur is not specified, the above reaction is assumed.*
**https://en.intl.chemicalaid.com/tools/equationbalancer.php?equation=H2S+%2B+SO2+%3D+S8+%2B+H2O

**Phases of `H_2S` & `H_2O` are assumed.

Given 2 moles `H_2S + 11.2L SO_2(g)` at STP
=> 2 `"moles"` `H_2S` + 0.50 mole `SO_2(g)` at STP
=> Limiting Reagent => 0.50 mole `SO_2(g)`; 1 mole `H_2S` excess

8 moles `SO_2(g)` => 3 moles `S_8(s)`
0.50 mole `SO_2(g)` => ? moles `S_8(s)`

`(8"mole"SO_2)/(0.50"mole"SO_2)` = `(3"mole"S_8(s))/(X)`

`X = ((0.50"mole"SO_2(g))(3"mole"S_8(s)))/(8"mole"SO_2(g))` = `0.188 "mole" S_8(s)`