Question

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Calculate the number of pounds of Co, released into the atmosphere when a 10.0 gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, CHg, and that the density of gasoline is 0.692 g - mL ! This assumption ignores additives. Also, assume complete combustion. CO, released

Answer 1

Reaction of gasoline with oxygen is 2C8H 18 + 2502 16602 + 18 H2O volume of tank of gasoline (W) = 1000 gallon 1 1 gallon = 3.785 L 00 Volumelu) = 10.0 X 3.785L = 37.854 Dersity of gasoline (d) = 0.692 g massim) .: Densityld) = volumelu) mass (m) = Density (d) x volume (V) mass of gasoline (m) = (0.692x1039) (37685L) g m= 26192.2g molar mass (mm) of gasoline = 114.2g|mol Mass " moles molas mass mass of gasoline 26192.29 So, Scanned wides of CamScanner gasoline - molas mass of gasoline 114.2g/mol

moles of gasoline = 229.35 mol 2C8H18 + 2502 - 16202 + 18 H20 By mol ratio (229.35 mol of gasoline) & ( 16 mol of (02) .= 1834, 8 mollo2 (2 mol of gasoline) moles of co2 = 1834.8 mol molal mass of co2= 44.01 9 Imol mass (602) = moles (602) x molal mass (602) = (44.01 ) (1834.8 moi) = 80749.sug mass of co2 released = 8074g.sug mass of (02= 8074g. 54 x103 kg. is 19= 10-3 kg) linkg = 2.204 tb) mass of co2 = 80749.54 X10-3x 2,2041 = 178.02 1b Scanned with CamScanne co2 released. 178.02 In