Give the masses of the three most abundant cl2o molecules

Chlorine has the following % abundance and mass:  

Cl-35 = 75.76 %, 34.9688 amu

Cl-37 = 24.24 %, 36.9659 amu

Oxygen has the following % abundance and mass:  

O-16 = 99.757 %, 15.9949 amu

O-17 = 0.038 %, 16.9991 amu

O-18 = 0.205 %, 17.9991 amu

Based in the % abundance O-16 is the most abundant oxygen isotope and Cl-35 is the most abundant chlorine isotope

So the most abundant molecule of Cl2O and their respective mass are as follows:  

1) Cl-35, Cl-35, O-16

Mass = 34.9688 amu + 34.9688 amu + 15.9949 amu = 85.9325 amu

2) Cl-35, Cl-37, O-16

Mass = 34.9688 amu + 36.9659 amu + 15.9949 amu = 87.9296 amu

3) Cl-37, Cl-37, O-16

Mass = 36.9659 amu + 36.9659 amu + 15.9949 amu = 89.9267 amu

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the dependant variable is spilled water

the independent is the heat