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

Answer:

The correct answer is option D.

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:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{1 atm\times 12 L}{0.0820 atm l /mol K\times 293.15 K}$

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:

$\frac{9}{2}\times 0.10 mol=0.45 mol$

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'

$V'=\frac{n'RT}{P}=\frac{0.0492 mol \times 0.0820 atm l /mol K\times 293.15 K}{1 atm}$

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 :

$\frac{6}{2}\times 0.10 mol=0.30 mol =n_c$

$V_c=\frac{n_cRT}{P}=\frac{0.30 mol \times 0.0820 atm l /mol K\times 293.15 K}{1 atm}=7.21 L$

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 :

$\frac{8}{2}\times 0.10 mol=0.40 mol =n_w$

$V_w=\frac{n_wRT}{P}=\frac{0.40 mol\times 0.0820 atm l /mol K\times 293.15 K}{1 atm}=9.61 L$

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