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

Jul 8, 2017

I think you refer to the so-called $\text{comproportionation reaction}$....

#### Explanation:

The sulfide in $\text{hydrogen sulfide}$ is OXIDIZED $\left(i\right)$ to elemental sulfur, and the $\text{sulfur dioxide}$ is REDUCED to elemental sulfur......$\left(i i\right)$

${S}^{2 -} \rightarrow S + 2 {e}^{-}$ $\left(i\right)$

$S {O}_{2} + 4 {H}^{+} + 4 {e}^{-} \rightarrow S + 2 {H}_{2} O$ $\left(i i\right)$

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 $2 \times \left(i\right) + \left(i i\right)$ to eliminate the electrons to give........

$2 {S}^{2 -} + S {O}_{2} + 4 {H}^{+} + \cancel{4 {e}^{-}} \rightarrow 3 S + 2 {H}_{2} O + \cancel{4 {e}^{-}}$

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

$2 {H}_{2} S + S {O}_{2} \rightarrow 3 S + 2 {H}_{2} O$

And now (finally) we can address your question. $\text{NTP}$ specifies a molar volume of $24.06 \cdot L \cdot m o {l}^{-} 1$......

With respect to $S {O}_{2}$ we have a molar quantity of $\frac{11.2 \cdot L}{24.06 \cdot L \cdot m o {l}^{-} 1} = 0.466 \cdot m o l \ldots .$. And thus $S {O}_{2}$ is the limiting reagent and $\text{hydrogen sulfide}$ is in excess.

And so we gets $3 \times$ an equivalent quantity of $S$, i.e. $\frac{3 \cdot S}{1 \cdot S {O}_{2}} \times 0.466 \cdot m o l s \times 32.06 \cdot g \cdot m o {l}^{-} 1 = 44.8 \cdot g$.

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

0.188 mole ${S}_{8} \left(s\right)$

#### Explanation:

$16 {H}_{2} S \left(g\right) + 8 S {O}_{2} \left(g\right)$ => $16 {H}_{2} O \left(l\right)$ + $3 {S}_{8} \left(s\right)$
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 ${H}_{2} S$ & ${H}_{2} O$ are assumed.

Given 2 moles ${H}_{2} S + 11.2 L S {O}_{2} \left(g\right)$ at STP
=> 2 $\text{moles}$ ${H}_{2} S$ + 0.50 mole $S {O}_{2} \left(g\right)$ at STP
=> Limiting Reagent => 0.50 mole $S {O}_{2} \left(g\right)$; 1 mole ${H}_{2} S$ excess

8 moles $S {O}_{2} \left(g\right)$ => 3 moles ${S}_{8} \left(s\right)$
0.50 mole $S {O}_{2} \left(g\right)$ => ? moles ${S}_{8} \left(s\right)$

$\left(8 \text{mole"SO_2)/(0.50"mole} S {O}_{2}\right)$ = $\frac{3 \text{mole} {S}_{8} \left(s\right)}{X}$

X = ((0.50"mole"SO_2(g))(3"mole"S_8(s)))/(8"mole"SO_2(g)) = $0.188 \text{mole} {S}_{8} \left(s\right)$