Question

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

the exothermic reaction, 620 L of CO2 was collected at 823 K and 750 torr. How
many liters of O2 gas at 540 torr and 298 K were consumed?
2 C,H10 (8) + 13 0,(8) →800,() + 10 H2O(g)

Answer 1

Answer: 510 L of oxygen gas were consumed.

Step-by-step 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=(PV)/(RT)`

`n=(0.99atm* 620L)/(0.0821L atm/K mol* 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 =
`(13)/(8)* 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=(nRT)/(P)`

`V=(14.8mol* 0.0821L atm/K mol* 298K)/(0.71atm)=510L`

Thus 510 L of oxygen gas were consumed.