Thank you in advanced! A Claus plant converts gaseous sulfur compounds to elemen

Question

Thank you in advanced!

A Claus plant converts gaseous sulfur compounds to elemental sulfur, thereby eliminating emission of sulfur into the atmosphere. The process can be especially important in the gasification of coal, which contains significant amounts of sulfur that is converted to H_2 S during gasification. In the Claus process, the H_2 S-rich product gas recovered from an acid-gas removal system following the gasifier is split, with one-third going to a furnace where the hydrogen sulfide is burned at 1 atm with a stoichiometric amount of air to form SO_2. H_2 + 3/2 O_2 rightarrow SO_2 + H_2 O The hot gases leave the furnace and are cooled prior to being mixed with the remainder of the H_2 S-rich gases. The mixed gas is then fed to a catalytic reactor where hydrogen sulfide and SO_2 react to form elemental sulfur. 2H_2 S + SO_2 rightarrow 2H_2 O + 3S The coal available to the gasification process is 0.6 wt% sulfur, and you may assume that all of the sulfur is converted to H_2 S, which is then fed to the Claus plant. Estimate the feed rate of air to the Claus plant in kg/kg coal. While the removal of sulfur emissions to the atmosphere is environmentally beneficial, identify an environmental concern that still must be addressed with the products from the Claus plant.

Explanation / Answer

(a): Given the S percentage of coal = 0.6 wt%

Let's consider 1 kg ( = 1000 g) of coal.

mass of S in 1Kg of coal = 1000 g x (0.6 / 100) = 6.0 g S

moles of S in 1 Kg of coal = mass / molecular mass = 6.0 g / 32.0 g/mol = 0.1875 mol S

Hence moles of H2S formed during gassification of coal = 0.1875 mol S x ( 1mol H2S / 1 mol S)

= 0.1875 mol H2S

Now 1/3 rd of H2S was treated with air.

Hence moles of H2S treated with air = 0.1875 mol H2S / 3 = 0.0625 mol S

The balanced chemical equation for the reaction of H2S with air is

H2S + 3/2O2 ---- > SO2 + H2O

1 mol, 3/2 mol, ---- 1 mol, 1 mol

1 mole of H2S reacts with the moles of O2 = 3/2 mol O2

Hence 0.0625 mol S that will react with the mole of O2 = 3/2 x 0.0625 = 0.09375 mol O2

Moles of air equivalent to 0.09375 mol O2 =  0.09375 mol O2 / 0.2 = 0.46875 mol air / Kg coal

Average molecular mass of air = 28.97 g/mol

Hence mass of air fed into the plant = (0.46875 mol air / Kg coal ) * 28.97 g/mol

= 13.58 g air / Kg coal

= 13.58 g * ( 1 Kg / 1000 g) air / Kg coal = 0.01358 Kg air / Kg coal (answer)

(b): Environmental concern caused by elemental sulphur is tarnishing of metals by elemetnal sulphur.

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