Answer: 510 L of oxygen gas were consumed.
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 750 torr = 0.99 atm  (760 torr= 1atm)
V = Volume of gas = 620 L
n = number of moles of carbon dioxide = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]823K[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{0.99atm\times 620L}{0.0821L atm/K mol\times 823K}=9.08moles[/tex]
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
According to stoichiometry:
8 moles of [tex]CO_2[/tex] are produced by = 13 moles of oxygen
Thus 9.08 moles of [tex]CO_2[/tex] are produced by = [tex]\frac{13}{8}\times 9.08=14.8[/tex] moles of oxygen
[tex]PV=nRT[/tex]
P = pressure of gas = 540 torr = 0.71 atm  (760 torr= 1atm)
V = Volume of gas = ?
n = number of moles of oxygen = 14.8
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]298K[/tex]
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{14.8mol\times 0.0821L atm/K mol\times 298K}{0.71atm}=510L[/tex]
Thus 510 L of oxygen gas were consumed.
