The complete combustion of octane, C8H18, the main component of gasoline, proceeds as shown below.
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)
(a) How many moles of O2 are needed to burn 1.00 mol of C8H18?
(b) How many grams of O2 are needed to burn 1.75 g of C8H18?
(c) Octane has a density of 0.692 g/mL at 20°C. How many grams of O2 are required to burn 3.35 gal of C8H18?
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(a) 25 moles of O2 are needed to burn 2 moles of C3H18. It is noted from the given reaction.
Hence, 12.5 moles of O2 will be needed to burn 1 mole of C3H18.
(b) 2C8H18 = 2(12*8+18*1) = 228
25O2 = 25*16*2 = 800
Hence, O2 required to burn 1.75 gm of C8H18 = (800/228)*1.75 = 6.14gms
(c) Weight of 3.35 gallon of octane = 3.35*1000*0.692ml*gm/ml = 2318.2gm
Hence O2 needed = (800/228)*2318.2 = 8134.04gms
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