The complete combustion of cotane,C 8 H 18 , the main component of gasoline,proc

Question

The complete combustion of cotane,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 1.40mol of C8H18?   _____ mol (b) How many grams of O2 are needed to burn 1.50 gof C8H18? _____ g (c) Octane has a density of 0.692 g/mL at 20 degrees C. How many grams of O2 are required to burn 4.40 gal ofC8H18?   _____ g The complete combustion of cotane,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 1.40mol of C8H18?   _____ mol (b) How many grams of O2 are needed to burn 1.50 gof C8H18? _____ g (c) Octane has a density of 0.692 g/mL at 20 degrees C. How many grams of O2 are required to burn 4.40 gal ofC8H18?   _____ g

Explanation / Answer

2 C8H18(l) + 25 O2(g) -->16 CO2(g) + 18 H2O(l) (a) How many moles of O2 are needed to burn 1.40mol of C8H18?   for 2 moles of octane it requires 25 moles of O2    1.4 moles of octane requires X moles of o2 X = ( 25*1.4 ) / 2     = 17.5 moles of O2 (b) How many grams of O2 are needed to burn 1.50 gof C8H18? Molar mass of C8H18 is = 8 * 12 + 18*1 = 114 g Molar mass of O2 = 2 * 16 = 32 g 2 * 114 g of octane requires 25* 32 g of O2 1.5 g of octane requires Y g of O2 Y = ( 25*32*1.5 ) / ( 2*114 )      = 5.2631 g of O2 (c) Octane has a density of 0.692 g/mL at 20 degrees C. How many grams of O2 are required to burn 4.40 gal ofC8H18? 1 gallon = 3785.4118 millilitres 4.4 gallon = 4.4 * 3785.4118 millilitres               = 16655.82 mL mass = density * Volume          = 0.692 g /mL * 16655.82 mL          = 11525.8 g 2 * 114 g of octane requires 25* 32 g of O2 11525.8 g ofoctane requires Z g of O2 Z = ( 25*32*11525.8 ) / ( 2*114 )    = 40441.48g of O2 2 * 114 g of octane requires 25* 32 g of O2 11525.8 g ofoctane requires Z g of O2 Z = ( 25*32*11525.8 ) / ( 2*114 )    = 40441.48g of O2

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