Question

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

Answer 1

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 2

Answer:The the correct answer is option (A).

Explanation:

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

12 L of oxygen gas burns 0.10 mol of propanol.(1 L = 1$dm^3$)

According to reaction, 2 moles of propanol gives 6 moles of $CO_2$ then, 0.10 moles of propanol will give:$\frac{6}{2}\times 0.10$ moles of $CO_2$ that is 0.30 mol.

$n_{CO_2}=0.30 mol$

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

$PV=nRT$

$PV=n_{CO_2}RT$

$V_{CO_2}=\frac{0.30 mol \times 0.0821 L atm/mol K\times 298 K}{1 atm}=7.33 L=7.33 dm^3$

The closest answer from the given options is option (A) that is 7.20 $dm^3$