Answer:
14.5g of CO₂ are produced
Explanation:
The reaction of octane with oxygen is:
C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O
<em>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.</em>
When 21.0 g of octane is burned with 19.0 g of oxygen gas you need to find <em>limiting reactant </em>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) = <em>2.3 moles of oxygen</em>
As you have just 0.594 moles of oxygen, <em>Oxygen is limiting reactant.</em>
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) =
<h3>14.5g of CO₂ are produced</h3>
