Question

When octane (C8H18) is burned in the presence of oxygen, the yield of products (carbon dioxide and water) is 87%. What mass of carbon dioxide will be produced in this engine when 21.0 g of octane is burned with 19.0 g of oxygen gas

Answer 1

Answer:

14.5g of CO₂ are produced

Explanation:

The reaction of octane with oxygen is:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

Where 1 mole of octane (Molar mass: 114.23g/mol) reacts with 25/2 moles of O₂ (Molar mass 32g/mol) to produce 8 moles of CO₂ and 9 moles of water.

When 21.0 g of octane is burned with 19.0 g of oxygen gas you need to find limiting reactant to find how many moles of products are formed:

Octane: 21.0g ₓ (1mol / 114.23g) = 0.184 moles octane

Oxygen: 19.0g ₓ (1 mol / 32g) = 0.594 moles oxygen

For a complete reaction of 0.184 moles of octane you will need:

0.184 moles C₈H₁₈ ₓ (25/2 moles O₂ / 1 mole C₈H₁₈) = 2.3 moles of oxygen

As you have just 0.594 moles of oxygen, Oxygen is limiting reactant.

Based on chemical equation, 25/2 of O₂ produce 8 moles of CO₂, that means theoretical yield of CO₂ with 0.594 moles of O₂ is:

0.594 moles O₂ ₓ (8 moles CO₂ / 25/2 moles O₂) = 0.380 moles of CO₂

But, as yield of products is 87%, moles produced of CO₂ are:

0.380 moles of CO₂ ₓ 87% = 0.331 moles CO₂ are produced.

As molar mass of CO₂ is 44g/mol, mass of CO₂ in 0.331 moles is:

0.331 moles CO₂ ₓ (44g / mol) =

14.5g of CO₂ are produced