The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows:
2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g)
| Relevant volumetric equivalencies |
|
1 gal = 3.785 L |
| 1 L = 1000 mL |
part C
Octane has a density of 0.692 g/mL at 20∘C. How many grams of O2 are required to burn 17.0 gal of C8H18?
Express the mass in grams to three significant digits.
!7 gal = 3.785 * 17 L = 3.785 * 1000 * 17 ml
= 64345 ml
mass = density * volume
= 0.692 g/ml * 64345 ml
= 44526.74 g
mole of octance = 44526.74 / 114 = 390.585 mol
as per reaction mole ration of oxygen to octance is 25 : 2
so, mole of oxygen needed = 12.5 * 380.585
= 4882.3125 mol
mass of oxygen = 4882.3125 * 32 = 156234 g
= 1.56 * 105 g
The complete combustion of octane, C8H18, a component of gasoline, proceeds as follows: 2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g) Relevant volumetric...
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