Respuesta :
Answer:
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= [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
∴ the maximum volume of ethanol (0.789 g/mL) that can be burned per minute = 0.895 mL
The maximum volume of ethanol that can be burnt per minute is 0.617 ml.
Based on the given information,
• The volume of air is 3.72 L, pressure is 1 atm, and given temperature is 25 degree C or 273 +25 = 298 K.
• The number of molecules of oxygen in air is 21%.
Now the formula for ideal gas equation is,
PV = nRT
The number of moles of air = PV/RT (R = 0.0821 Latm/mol/K)
[tex]nair = \frac{1*3.72}{0.0821*298} \\nair = 0.15[/tex]
As the number of molecules of oxygen in air is 21%, now the number of moles of oxygen in air will be,
[tex]= \frac{21*0.15}{100} \\= 0.0317[/tex]
Now, the reaction of the burning of ethanol is,
C₂H₅OH + 3O₂ ⇔ 2CO₂ +3H₂O + heat
• 1 mole of C₂H₅OH requires 3 moles of oxygen to burn completely or it can be said that 1 mole of oxygen burns 1/3 moles of C₂H₅OH.
• Therefore, 0.0317 moles of O₂ burns = 0.0317/3 or 0.0106 moles of C₂H₅OH.
Now the molar mass of C₂H₅OH is 46 g/mol, and the mass of C₂H₅OH will be,
[tex]Mass of ethanol = No of moles * Molar mass\\Mass of ethanol = 0.0106 * 46 g/mol\\Mass of ethanol = 0.487 grams\\[/tex]
Now the density of C₂H₅OH is 0.789 g/ml, the volume can be determined by using the formula,
Volume = Mass/Density
[tex]Volume = \frac{0.487 g}{0.789 g/ml} \\Volume = 0.617 ml[/tex]
Thus, the maximum volume of ethanol that can be burnt per minute is 0.617 ml.
To know more about:
https://brainly.com/question/13341164
