Answer:
16.5g of CO₂ could be produced
Explanation:
The combustion of hexane occurs as follows:
C₆H₁₄(l) + 19/2O₂ → 6CO₂ + 7H₂O
Where 1 mole of hexane reacts with 19/2 moles of O₂.
To solve this question we need to find the moles of each reactant in order to find limiting reactant. The moles of the limiting reactant will determine the moles of CO₂ produced:
Moles C₆H₁₄ -Molar mass: 86.18g/mol-:
17.2g hexane * (1mol / 86.18g) = 0.200 moles hexane
Moles O₂ -Molar mass: 32g/mol-:
19g O₂ * (1mol / 32g) = 0.594 moles oxygen.
For a complete reaction of 0.594 moles of oxygen are required:
0.594 moles O₂ * (1mol C₆H₁₄ / 19/2 moles O₂) = 0.0625 moles C₆H₁₄.
As there are 0.200 moles of hexane, hexane is the excess reactant and oxygen the limiting reactant.
The moles of CO₂ produced assuming a yield of 100% -All moles of oxygen react producing carbon dioxide.:
0.594 moles O₂ * (6mol CO₂ / 19/2 moles O₂) = 0.375 moles of CO₂ could be produced. The mass is:
0.375 moles of CO₂ * (44.01g / mol) =
