Answer to Question #71427 in General Chemistry for A.Cohen

Question

how many grams of oxygen are necessary to drive a car 250 miles if your fuel mileage is 35 miles per gallon and you are using ethanol as your fuel?

Answer 1

C2H5OH + 3O2  2CO2 + 3H2O
1 mole of ethanol reacts with 3 moles of oxygen.
V = 250/35 = 7.143 (gal) – volume of fuel needed;
English gallon – 4.5461 L
American gallon – 3.785 L
Solution for English gallon:
V = 4.5461 × 7.143 = 32.473 (L) = 32 473 (mL) – volume of fuel needed;
m = 32 473 × 0.7893 = 25 631 (g) – mass of ethanol needed;
ν = (25 631)/46 = 557.2 (mole) – amount of ethanol needed;
ν = 557.2 × 3 = 1671.6 (mole) – amount of oxygen needed;
m = 1671.6 × 32 = 53 491.2 (g) – of oxygen needed.
Solution for American gallon:
V = 3.785 × 7.143 = 27.036 (L) = 27 036 (mL) – volume of fuel needed;
m = 27 036 × 0.7893 = 21 340 (g) – mass of ethanol needed;
ν = (21 340)/46 = 463.9 (mole) – amount of ethanol needed;
ν = 463.9 × 3 = 1391.7 (mole) – amount of oxygen needed;
m = 1391.7 × 32 = 44 534.4 (g) – of oxygen needed.
Answer
53 491.2 (g) of oxygen are necessary to drive a car 250 miles if your fuel mileage is 35 miles per English gallon.
44 5344 (g) of oxygen are necessary to drive a car 250 miles if your fuel mileage is 35 miles per American gallon.