The combustion of fuel in your car engine requires oxygen gas, which is supplied as air (21% oxygen molecules) into the engine. Consider a car that is using 100% ethanol, C2H5OH, as fuel. If your engine intakes 3.72 L of air per minute at 1.00 atm and 25ºC, what is the maximum volume of ethanol (0.789 g/mL) that can be burned per minute?

0.895 mL

Explanation:

The equation for the reaction can be written as:

C₂H₅OH + 3 O₂ ⇒ 2 CO₂ + 3 H₂O

Using ideal gas law

PV = nRT

where;

Pressure (P) = 1 atm

Volume (V) =  (3.2 L of air per minute)

General gas constant (R) =  0.082 atmL/molK

Temperature (T)=  25°C = 298.15 K

number of moles (n) = 0.192 mole

number of moles in 21% oxygen molecule will contain:

0.192 × 0.21 = 0.04032 moles

0.04032 moles O₂=

= 0.895 mL of ethanol per minute

∴ the maximum volume of ethanol (0.789 g/mL) that can be burned per minute = 0.895 mL