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)
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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