Question

Suppose we have a 12.2 L sample containing 0.50 mol oxygen gas at a pressure of 1 atm and a temperature of 25 degrees celcius. If all this oxygen were converted to ozone at the same temperature and pressure, what would be the volume of the ozone?Balanced equation: 3O2 → 2O3a. 6.5Lb. 2.8Lc. 8.1Ld. 7.4L

Answer 1

c. 8.1 L

Step-by-step explanation:

Given that:-

Moles of oxygen gas = 0.50 mol

According to the reaction shown below as:-

`3O_2\rightarrow 2O_3`

3 moles of oxygen gas on reaction gives 2 moles of ozone

Also,

1 mole of oxygen gas on reaction gives 2/3 moles of ozone

So,

0.50 mole of oxygen gas on reaction gives
`(2)/(3)* 0.50` moles of ozone

Moles of ozone = 0.3333 mol

Pressure = 1 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (25.0 + 273.15) K = 298.15 K

Volume = ?

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K

⇒V = 8.1 L