For the reaction
? s+? 02 →? s03,
what is the maximum amount of so3
(80.0632 g/mol) which could be formed from
17.9 g of s (32.065 g/mol) and 0.157 mol of
02 (31.9988 g/mol)?
answer in units of mol.

0.558mole of SO₃

Explanation:

Given parameters:

Molar mass of SO₃ = 80.0632g/mol

Mass of S = 17.9g

Molar mass of S = 32.065g/mol

Number of moles of O₂ = 0.157mole

Molar mass of O₂ = 31.9988g/mol

Unknown:

Maximum amount of SO₃

Solution

  We need to write the proper reaction equation.

           2S + 3O₂ → 2SO₃

We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.

So we simply compare the molar relationship between sulfur and product formed to solve the problem:

First, find the number of moles of Sulfur, S:

   Number of moles of S =

   Number of moles of S =   = 0.558mole

Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:

       2 mole of Sulfur produced 2 mole of SO₃

   Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃

a corroding iron, if that's one of your choices