The chemical reaction is
C3H8O + (9/2) O2 = 3CO2 + 4H2O
Theoretical amount of oxygen needed = 0.1 mol of propanol ( 4.5 mol O2 / 1 mol propanol) = 0.45 mol
The amount of oxygen used in mol can be calculated using the ideal gas equation
PV = nRT
@room conditions
P = 1 atm
T = 25C ~ 298 K
R = 0.08205 mol – L/ atm – K
1 dm3 = 1 L
(1 atm) (12 L) = n (0.08205)*(298K)
n = 0.49 mol O2
Excess O2 = 0.49 – 0.45 = 0.04 mol O2
Amount of CO2 produced = 0.1 mol propanol (3 mol CO2/ 1 mol propanol) = 0.3 mol
Amount of H2) produced = 0.1 mol propanol (4 mol H2O/ 1 mol propanol) = 0.4 mol
TOTAL amount of gases = 0.04 + 0.3 + 0.4 = 0.74 mol
Therefore
V = (0.74*0.08205*298)/1
V = 18.0 dm3, the answer is letter D
Answer:The the correct answer is option (A).
Explanation:
[tex]2C_3H_7OH+9O_2\rightarrow 6CO_2+8H_2O[/tex]
12 L of oxygen gas burns 0.10 mol of propanol.(1 L = 1[tex]dm^3[/tex])
According to reaction, 2 moles of propanol gives 6 moles of [tex]CO_2[/tex] then, 0.10 moles of propanol will give:[tex]\frac{6}{2}\times 0.10 [/tex] moles of [tex]CO_2[/tex] that is 0.30 mol.
[tex]n_{CO_2}=0.30 mol[/tex]
The final volume of the gas evolved after the reaction can be determined by Ideal gas equation. Since, the reaction is taking place at room temperature the value of pressure and temperature will be:
At room temperature:
Pressure = 1 atm
Temperature , T= 293 K
[tex]PV=nRT[/tex]
[tex]PV=n_{CO_2}RT[/tex]
[tex]V_{CO_2}=\frac{0.30 mol \times 0.0821 L atm/mol K\times 298 K}{1 atm}=7.33 L=7.33 dm^3[/tex]
The closest answer from the given options is option (A) that is 7.20 [tex]dm^3[/tex]
