Question

Butane and oxygen were allowed to completely react at 540 torr and 298 K. After

Answer 1

Answer: 510 L of oxygen gas were consumed.

Explanation:

According to ideal gas equation:

$PV=nRT$

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 =$0.0821Latm/Kmol$

T =temperature =$823K$

$n=\frac{PV}{RT}$

$n=\frac{0.99atm\times 620L}{0.0821L atm/K mol\times 823K}=9.08moles$

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

According to stoichiometry:

8 moles of $CO_2$ are produced by = 13 moles of oxygen

Thus 9.08 moles of $CO_2$ are produced by = $\frac{13}{8}\times 9.08=14.8$ moles of oxygen

$PV=nRT$

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 =$0.0821Latm/Kmol$

T =temperature =$298K$

$V=\frac{nRT}{P}$

$V=\frac{14.8mol\times 0.0821L atm/K mol\times 298K}{0.71atm}=510L$

Thus 510 L of oxygen gas were consumed.