Answer:
The limiting reagent is the O₂
Explanation:
We can think, this reaction
2O₂(g) + H₂(g) + S(s) → H₂SO₄
Mole of each = Mass / molar mass
6 g / 32 g/m = 0.187 mole O₂
4g / 2 g/m = 2 mole H₂
5g / 32.06 g/m = 0.156 mole S
Ratio between reactants is 2:1:1, 1:2:1, 1:1:2
For 2 mole of O₂, I need to react 1 mol of H₂ and 1 mol of S
0.187 mole of O₂, I need (the half)
0.093 mole of H₂ and 0.093 mole of S
For 1 mole of H₂, I need to react 2 mole of O₂ and 1 mol of S
2 mole of H₂, I need (the double of O₂ and the same for S)
4 mole of O₂ ; 2 mole of S
For 1 mol of S, I need to react 1 mol of H₂ and 2 mole of O₂
0.156 mole I need the same amount for H₂ and the double for O₂
0.156 mole of H₂ and 0.312 mole of O₂
In both cases, I can't make react, all the mass of oxygen, so this is the limiting reagent.
