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 = [tex]V=12.0 dm^3[/tex] = 12.0 L
Using Ideal gas equation:
[tex]PV=nRT[/tex]
[tex]n=\frac{PV}{RT}=\frac{1 atm\times 12 L}{0.0820 atm l /mol K\times 293.15 K}[/tex]
n = 0.4992033 moles of oxygen gas.
Initial number of moles of oxygen gas = 0.4992 mol
[tex]2C_3H_7OH+9O_2\rightarrow 6CO_2+8H_2O[/tex]
According to reaction, 2 moles of propanol reacts with 9 moles of oxygen.
Then 0.10 moles of propanol will react with:
[tex]\frac{9}{2}\times 0.10 mol=0.45 mol[/tex]
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'
[tex]V'=\frac{n'RT}{P}=\frac{0.0492 mol \times 0.0820 atm l /mol K\times 293.15 K}{1 atm}[/tex]
V' = 1.18 L
Volume of the [tex]CO_2 gas [/tex]
According to reaction, 2 moles of propanol gives with 6 moles of [tex]CO_2[/tex].
Then 0.10 moles of propanol will give :
[tex]\frac{6}{2}\times 0.10 mol=0.30 mol =n_c[/tex]
[tex]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[/tex]
Volume of the [tex]H_2O gas [/tex]
According to reaction, 2 moles of propanol gives with 8 moles of [tex]H_2O[/tex].
Then 0.10 moles of propanol will give :
[tex]\frac{8}{2}\times 0.10 mol=0.40 mol =n_w[/tex]
[tex]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[/tex]
Total volume of the gas measured = [tex]V'+V_c+V_w=1.81 l+7.21 L+9.61 L=17.97 L\approx 18 L[/tex]
