Question

CO2 will be produced when 1.60 mol of CS2 burns with 5.60 mol of O2. Determine the limiting reactant

Answer 1

Solution:

The balanced chemical equation:

CS₂ + 3O₂ → CO₂ + 2SO₂

According to the equation above: n(CS₂) = n(O₂)/3

Moles of CS₂ = 1.60 mol

Moles of O₂ = 5.60 mol

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose CS₂:

n(O₂) = 3 × n(CS₂) = 3 × 1.60 mol = 4.80 mol

The calculation above means that we need 4.80 mol of O₂ to completely react with CS₂.

We have 5.60 mol of O₂ and therefore more than enough oxygen.

Thus oxygen (O₂) is in excess and carbon disulfide (CS₂) must be the limiting reagent.

Hence, the limiting reactant is carbon disulfide (CS₂).

Answer: The limiting reactant is CS₂.