Answer:
24.6g of CO₂ is theoretical yield
Step-by-step explanation:
The reaction of 8.00g of octane with 38.9g of oxygen.
The reaction of octane with oxygen is:
C₈H₁₈(l) + 25/2O₂ → 9H₂O + 8CO₂
1 mole of octane reacts with 25/2 moles of oxygen to produce 8 moles of CO₂
Theoretical yield is the amount of carbon dioxide formed assuming a yield of 100%. To calculate theoretical yield, first, we need to find limiting reactant and, with the chemical reaction, we can obtain the theoretical moles of CO₂ produced and its mass to obtain theoretical yield.
Limiting reactant:
Moles octane (Molar mass: 114.23g/mol) in 8.00g:
8.00g × (1mol / 114.23g) = 0.0700 moles octane.
Moles oxygen (Molar mass: 32g/mol) in 38.9g:
38.9g × (1mol / 32g) = 1.2156 moles oxygen.
For a complete reaction of 1.2156 moles of O₂ there are necessaries:
1.2156 moles O₂ ₓ (1mol C₈H₁₈ / 25/2 moles O₂) = 0.0973 moles octane
As we have just 0.0700 moles,
octane is limiting reactant.
Moles and mass of carbon dioxide:
As limiting reactant is octane, 0.0700 moles of C₈H₁₈ will produce:
0.0700mol C₈H₁₈ × (8 moles CO₂ / 1 mol C₈H₁₈) = 0.56 moles of CO₂ are theoretically produced. In mass (Molar mass CO₂ = 44.01g/mol):
0.56moles CO₂ × (44.01g / mol) =
24.6g of CO₂ is theoretical yield
-Theoretical yield because we are assuming all octane is reacting. In real life, never happens like that-