Unit 7 A new liquid fuel, which has a molar mass of 112 g/mol and a density of 0
ID: 541846 • Letter: U
Question
Explanation / Answer
a)
V = 20.6 gal --> change to L
V = 20.6*3.785 = 77.971 L = 77.971*10^3 mL
change to mass
M = D*V = (77.971*10^3)(0.712) = 55515.352 g/tank
mol = mass/MW = 55515.352/112 = 495.6727 mol/tank
E = n*HRxn = (495.6727)(5.55*10^3) = 2750983.485 kJ/tank
t = E/rate = (2750983.485)/(5.62*10^4) = 48.949 h
D = v*t = (60 mi/h)(48.949 h) = 2936.94 mi
D = 2936.94 mi * 1.6 km/mi = 4726.54 km
therefore, 1 tank = 4726.54 km
b)
if actual rang eis 732 km V = 60 mph
find efficeincy
% Eff = real / theoretical * 100% = 732 / 4726.54 *100 = 15.48%
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