Answer:
4.51 × 10³ kcal
B. less
Explanation:
Let's consider the combustion of 2-methylheptane.
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
When 1 mole of C₈H₁₈ burns, 1.306 × 10³ Kcal are released (heat of combustion). The molar mass of C₈H₁₈ is 114 g/mol. Then, for 394 g of C₈H₁₈:
[tex]394gC_{8}H_{18}.\frac{1molC_{8}H_{18}}{114gC_{8}H_{18}}.\frac{1.306 \times 10^{3} kcal }{1molC_{8}H_{18}} =4.51 \times 10^{3} kcal[/tex]
Assuming the same efficiency, would 394 grams of ethanol provide more, less, or the same amount of energy as 394 grams of 2-methylheptane?
Let's consider the combustion of ethanol.
C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O
When 1 mole of C₂H₆O burns, 326.7 Kcal are released (heat of combustion). The molar mass of C₂H₆O is 46.0 g/mol. Then, for 394 g of C₂H₆O:
[tex]394gC_{2}H_{6}O.\frac{1molC_{2}H_{6}O}{46.0gC_{2}H_{6}O}.\frac{326.7 kcal }{1molC_{2}H_{6}O} =2.80 \times 10^{3} kcal[/tex]
The combustion of 394 g of ethanol provides less energy than the combustion of 394 g of 2-methylheptane.
