Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
PV = nRT ; n = PV/RT
Substituting the known values,
n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
volume = 0.0833 L or 83.34 mL
Answer:
A)
Cl + O3 --> ClO + O2
2ClO --> ClOOCl
ClOOCl --> 2 Cl + O2
B)
Cl2 + 4O3 -> 2 ClO + 5O2
Explanation:
Step 1: Chlorine atoms react with ozone (O3) to form chlorine monoxide and molecular oxygen
Cl + O3 --> ClO + O2
Step 2: Chlorine monoxide molecules combine to form ClOOCl gas
2ClO --> ClOOCl
Step 3: ClOOCl absorbs sunlight and breaks into chlorine atoms and molecular oxygen
ClOOCl --> 2 Cl + O2
B) Overall Balanced equation
Cl2 + 4O3 -> 2 ClO + 5O2
For every 2 mol of Na 1 mole of cl2 will react
4 mol Na = 2mole cl2
