Question

What volume of oxygen gas measured at 0 degree Celsius and 1 atm is needed to burn completely 1 litre of propane gas measured under the same conditions?

Answer 1

Answer:

Explanation:

C3H8 + 5O2----- 3CO2+ 4H2O

1 mole of C3H8 we have

5 mole of O2 we have

From balanced eqn..

Mole=n=v/22.4

For propane ,1= v/22.4

Or v =22.4

For oxygen,5= v/22.4

Or v=5*22.4

Vol of O2/vol of propane

5*22.4/22.4

=5 LTRs.

Hope this helps

Answer 2

Answer: 5 liters

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = 1 L

T= Temperature of the gas = 0°C = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= ?

$n=\frac{PV}{RT}=\frac{1\times 1}0.0821\times 273}=0.04moles$

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

According to stoichiometry :

1 mole of propane requires 5 moles of oxygen

Thus 0.04 moles of propane require=$\frac{5}{1}\times 0.04=0.22moles$ moles of oxygen

Volume of oxygen $moles\times {\text {molar volume}}=0.22\times 22.4L=5L$

Thus 5 Liters of oxygen gas  is needed to burn completely 1 litre of propane gas measured under the same conditions.