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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.

Answer :

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Answer:

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 ×[tex]\frac{21 molesO_2}{100 molesAIR}[/tex] = 0,0406 moles O₂

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

0,046 moles O₂[tex]\frac{1molC_{2}H_{5}OH}{3molesO_2} \frac{46,07 g}{1mol} \frac{1mL}{0,789g}[/tex] = 0,895 mL of ethanol per minute

I hope it helps!

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