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

2 Answers

Answer:

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......

Jul 8, 2017

Answer:

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)#