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

c. 8.1 L

Explanation:

Given that:-

Moles of oxygen gas = 0.50 mol

According to the reaction shown below as:-

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 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

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.