Answer to Question #169353 in Chemistry for Emily

Question

How many grams of CO2 will be produced when 7.60mol of CS2 burns with 5.90 mol of O2?

Answer 1

Solution:

The balanced chemical equation:

CS₂ + 3O₂ → CO₂ + 2SO₂

According to the equation above: n(CS₂) = n(O₂)/3 = n(CO₂)

Moles of CS₂ = 7.60 mol

Moles of O₂ = 5.90 mol

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose O₂:

n(CS₂) = n(O₂) / 2 = 5.90 mol / 2 = 1.9667 mol​

The calculation above means that we need 1.9667 mol of CS₂ to completely react with O₂.

We have 7.60 mol of CS₂ and therefore more than enough carbon disulfide.

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

Thus: n(CO₂) = n(O₂)/3 = 5.90 mol / 3 = 1.9667 mol

Moles of CO₂ = Mass of CO₂ / Molar mass of CO₂

Mass of CO₂ = Moles of CO₂ × Molar mass of CO₂

The molar mass of CO₂ is 44.01 g mol^-1.

Hence,

Mass of CO₂ = 1.9667 mol × 44.01 g mol^-1 = 86.553 g = 86.55 g

Mass of CO₂ = 86.55 g

Answer: 86.55 grams of CO₂ will be produced.