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 4.73 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? hint: you can ignore the "per minute" information because both the ethanol and air are being quantified per minute. enter your answer to three significant figures in units of ml.

the maximum volume that can be burned per minute is: 0,895 mL of ethanol.

Explanation:

The combustion of ethanol is:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

With gas law:

PV/RT = n

Where P is pressure (1,00 atm)

V is volume (4,73 L of air per minute)

R is gas constant (0,082 atmL/molK)

T is temperatue(25°C≡298,15K)

And n are moles, replacing:

n = 0,193 moles of air per minute.

These moles of air contain:

0,193 moles air × = 0,0406 moles O₂

Thus, the maximum volume that can be burned per minute is:

0,046 moles O₂ = 0,895 mL of ethanol per minute

I hope it helps!

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.