if 0.479 mol of potassium sulfide are reacted with 0.195 mol of aluminum sulfate, how many moles of aluminum sulfide will be produced?

3K₂S(aq) + Al₂(SO₄)₃(aq) → 3K₂SO₄(aq) + Al₂S₃(s)
Mole ratio K₂S : Al₂(SO₄)₃ = 3 : 1

When 0.479 mol of K₂S completely reacts, Al₂(SO₄)₃ needed = (0.479 mol) × (1/3) = 0.160 mol < 0.195 mol
Hence, Al₂(SO₄)₃ is in excess, and K₂S completely reacts.

According to the above equation, mole ratio K₂S : Al₂S₃ = 1 : 3
No. of moles of K₂S (completely) reacted = 0.479 mol
No. of moles of Al₂S₃ produced = (0.479 mol) × (1/3) = 0.160 mol