Question

100 points! please help me will mark brainliest

Answer 1

S + O2 → SO2

Moles of S:

11/32.06 = 0.3431 mol

Moles of O2:

44/15.999 = 2.75 mol

Sulfur will be the limiting reagent.

Oxygen will be the excess reagent

2.75 - 0.3431 = 2.4071 mols of oxygen leftover

2.4071 * 15.999 = 38.511

38.511 grams of oxygen will be leftover

Answer 2

Reaction

S + O₂ ⇒ SO₂

mole S = 11 : 32 g/mol = 0.34375

mole O₂ = 44 : 32 g/mol = 1.375

S as a limiting reactant (smaller) and converted all to sulfur dioxide

mole O₂ (unreacted) = 1.375 - 0.34375 = 1.03125

mass O₂ (unreacted) = 1.03125 x 32 = 33 g