Question

One of the steps to sweeten sour gas using the Claus process is reacting

hydrogen sulfide gas with sulfur dioxide gas to produce water vapour and sulfur.
16 H2S(g) + 8 SO2(g) → 16 H2O(g) + 3 Sg(s)
8.56 kL of hydrogen sulfide at 175 kPa and 250 °C reacts with excess sulfur
dioxide. Calculate the mass, in kg, of sulfur produced.

Answer 1

The mass, in kg, of Sulfur produced : 2.072 kg

Further explanation

Given

V = 8.56 kL = 8560 L

P = 175 kPa = 1,73 atm

T = 250 + 273 = 523 K

Required

mass of Sulfur produced

Solution

mol of H₂S :

`\tt n=(PV)/(RT)\\\\n=(1.73* 8560)/(0.082* 523)\\\\n=345.3`

mol of Sulfur based on mol H₂S as a limiting reactant( excess Sulfur dioxide)

From equation, mol ratio H₂S : S = 16 : 3, so mol S :

`\tt (3)/(16)* 345.3=64.74`

Mass S(Ar = 32 g/mol) :

= mol x Ar s

= 64.74 x 32

= 2071.68 g = 2.072 kg