For the reaction 2 s (s) + 3 o2 (g) → 2 so3 (g), how much so3 can be produced from 4 g o2 and excess s? 1. 0.25 mol so3 2. 0.13 mol so3 3. 0.06 mol so3 4. 0.08 mol so3 5. 0.17 mol so3

The balanced equation for the above reaction is as follows;

2S + 3O₂ - - > 2SO₃

Stoichiometry of O₂ to SO₃ is 3:2

O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.

Number of O₂ moles reacted - 4 g / 32 g/mol = 0.125 mol

3 mol of O₂ forms 2 mol of SO₃

therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol

Number of SO₃ moles formed - 0.0833 mol

Answer is 4) 0.08 mol