Problem: You may want to reference (Pages 102 - 105)Section 3.6 while completing this problem.The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows: 2 C8H18(l) + 25 O2(g)  →  16 CO2(g) + 18 H2O(g)Relevant volumetric equivalencies1 gal = 3.785 L1 L = 1000 mLOctane has a density of 0.692 g/mL at 20oC. How many grams of O2 are required to burn 17.0 gal of C8H18?

FREE Expert Solution

The balanced reaction is:

2 C8H18(l) + 25 O2(g)  →  16 CO2(g) + 18 H2O(g)

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Problem Details
You may want to reference (Pages 102 - 105)Section 3.6 while completing this problem.
The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows: 2 C8H18(l) + 25 O2(g)  →  16 CO2(g) + 18 H2O(g)
Relevant volumetric equivalencies

1 gal = 3.785 L

1 L = 1000 mL
Octane has a density of 0.692 g/mL at 20oC. How many grams of O2 are required to burn 17.0 gal of C8H18?

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