The Claus reactions, shown below, are used to generate elemental sulfur from hydrogen sulfide. 2H2S + 3O2 → 2SO2 + 2H2O SO2 + 2H2S → 3S + 2H2OHow much sulfur (in grams) is produced from 23.0 grams of O2?



Answer :

Explanation:

The Claus reactions are:

2 H₂S + 3 O₂ -----> 2 SO₂ + 2 H₂O

SO₂ + 2 H₂S -----> 3 S + 2 H₂O

First the mass of O₂ must be converted into moles using the molar mass of O₂.

atomic mass of O = 16.00 amu

molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32.00 g/mol

moles of O₂ = mass of O₂/molar mass of O₂

moles of O₂ = 23.0 g/32.00 g/mol

moles of O₂ = 0.719 moles

According to the first reaction, 2 moles of H₂S will react with 3 moles of O₂ to give 2 moles of SO₂ and 2 moles of H₂O. Then, the mole ratio between O₂ and SO₂ is 3 to 2. That relationship can be used to get the number of moles of SO₂ produced by 0.719 moles of O₂. We select SO₂ because it is one of the reactants of the second reaction and it can link both reactions.

2 moles of SO₂ = 3 moles of O₂

moles of SO₂ = 0.719 moles of O₂ * 2 moles of SO₂/(3 moles of O₂)

moles of SO₂ = 0.479 moles

According to the second equation 3 moles of S can be produced by 1 mol of SO₂. That relationship can be used to find the number of moles of S produceeb by 0.479 moles of SO₂.

1 mol of SO₂ = 3 moles of S

moles of S = 0.479 moles of SO₂ * 3 moles of S/(1 mol of SO₂)

moles of S = 1.44 moles

Finally the moles of S produced can be converted back to grams using the molar mass of S.

atomic mass of S = 32.07 amu

molar mass of S = 32.07 g/mol

mass of S = moles of S * molar mass of S

mass of S = 1.44 moles * 32.07 g/mol

mass of S = 46.2 g

Answer: 46.2 g of sulfur is produced from 23.0 grams of oxygen.