Question

In an experiment, 12.0dm3 of oxygen, measured under room conditions, is used to burn completely 0.10mol of propan-1-ol. What is the final volume of gas, measured under room conditions? A 7.20dm3 B 8.40dm3 C 16.8dm3 D 18.00dm3 ans is B. Pls explain how

Answer 1

2C₃H₇OH + 9O₂ = 6CO₂ + 8H₂O

V(O₂)=12.0 dm³
n(C₃H₇OH)=0.1 mol

n(O₂)=12.0 dm³/22.4 dm³/mol=0.5357 mol

C₃H₇OH : O₂ 2:9 1:4.5

0.1:0.5357
oxygen in excess

V(CO₂)=3Vm*n(C₃H₇OH)

V(CO₂)=3*22.4*0.1=6.72 dm³

Answer 2

The correct answer is option D.

Step-by-step explanation:

At room conditions ,temperature is taken as T =293.15 k

At room conditions, pressure is taken as P = 1 atm

Given the volume of oxygen gas =
`V=12.0 dm^3` = 12.0 L

Using Ideal gas equation:

`PV=nRT`

`n=(PV)/(RT)=(1 atm* 12 L)/(0.0820 atm l /mol K* 293.15 K)`

n = 0.4992033 moles of oxygen gas.

Initial number of moles of oxygen gas = 0.4992 mol

`2C_3H_7OH+9O_2\rightarrow 6CO_2+8H_2O`

According to reaction, 2 moles of propanol reacts with 9 moles of oxygen.

Then 0.10 moles of propanol will react with:

`(9)/(2)* 0.10 mol=0.45 mol`

Final moles of oxygen gas left after combustion reaction = n'

n'= 0.4992 mol - 0.45 mol = 0.0492 mol

So, the final volume of the oxygen gas measured will be given as:V'

`V'=(n'RT)/(P)=(0.0492 mol * 0.0820 atm l /mol K* 293.15 K)/(1 atm)`

V' = 1.18 L

Volume of the
`CO_2 gas`

According to reaction, 2 moles of propanol gives with 6 moles of
`CO_2`.

Then 0.10 moles of propanol will give :

`(6)/(2)* 0.10 mol=0.30 mol =n_c`

`V_c=(n_cRT)/(P)=(0.30 mol * 0.0820 atm l /mol K* 293.15 K)/(1 atm)=7.21 L`

Volume of the
`H_2O gas`

According to reaction, 2 moles of propanol gives with 8 moles of
`H_2O`.

Then 0.10 moles of propanol will give :

`(8)/(2)* 0.10 mol=0.40 mol =n_w`

`V_w=(n_wRT)/(P)=(0.40 mol* 0.0820 atm l /mol K* 293.15 K)/(1 atm)=9.61 L`

Total volume of the gas measured =
`V'+V_c+V_w=1.81 l+7.21 L+9.61 L=17.97 L\approx 18 L`