Consider the reaction 2 s(s) + 3 o2(g) → 2 so3(g). how many moles of so3 will be produced from 4.0 mol o2 and excess s?

The moles of sulfur trioxide produced are 2.67 moles.

Explanation:

We are given:

Moles of oxygen gas = 4.0 moles

The given chemical equation follows:

Solid sulfur is present in excess. So, it is considered as an excess reagent. And, oxygen gas is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of sulfur trioxide

So, 4.0 moles of oxygen gas produces = of sulfur trioxide

Hence, the moles of sulfur trioxide produced are 2.67 moles.