3.12 mol of S and 3.75 mol of O2 react according to the equation 2S (s) + 3O2 (g) → 2SO3 (g). Identify the limiting reactant. A) S

B) About 2

C) O

D) SO 3

O₂

Explanation:

2S(s) +3O₂(g) → 2SO₃(g)

The limiting reactant of a chemical reaction is one that we do not have enough of for the reaction to yield more products.

From the balanced equation above, the mole ratio of S to O₂ is 2 :3. This means that for every 2 moles of S, 3 moles of O₂ is needed.

Since we have 3.12 mol of S,

amount of O₂ needed

= (3.12 ÷2) ×3

= 4.68 mol

Since we only have 3.75 mol of O₂, O₂ is the limiting reactant.

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