Ozone is depleted in the stratosphere by chlorine from CF3Cl according to the following set of equations:
CF3Cl + UV Light ---> CF3 + Cl
Cl + O3 ---> ClO + O2
O3 + UV Light ---> O2 + O
ClO + O ----> Cl + O2
What total volume of ozone measured at a pressure of 25.0 mmHg and a temperature of 235 K can be destroyed when all of the chlorine from 15.5g of CF3Cl goes through ten cycles of the above reactions?
how many moles of CF3Cl are in 15.5 grams?
I appears to me that two moles of ozone per cycle are destroyed per one mole of CF3Cl
0.005
To find the total volume of ozone destroyed, we need to calculate the moles of CF3Cl reacted and use stoichiometry to find the moles of O3 destroyed in each cycle.
First, let's find the moles of CF3Cl using its molar mass.
Molar mass of CF3Cl = (3 x atomic mass of F) + atomic mass of Cl
= (3 x 19.0 g/mol) + 35.5 g/mol
= 57.0 g/mol + 35.5 g/mol
= 92.5 g/mol
Now we can calculate the moles of CF3Cl reacted:
Moles of CF3Cl = Mass of CF3Cl / Molar mass of CF3Cl
= 15.5g / 92.5 g/mol
= 0.1676 mol
Next, we need to determine the moles of O3 destroyed in one cycle. Let's examine the stoichiometry of the reactions:
1 mole of CF3Cl reacts with 1 mole of O3
Therefore, the number of moles of O3 destroyed in 1 cycle is equal to the moles of CF3Cl reacted:
Moles of O3 destroyed in 1 cycle = Moles of CF3Cl
= 0.1676 mol
Since there are 10 cycles, we can calculate the moles of O3 destroyed in all 10 cycles:
Moles of O3 destroyed (10 cycles) = Moles of O3 destroyed in 1 cycle x 10
= 0.1676 mol x 10
= 1.676 mol
Now we can use the ideal gas law to find the volume of ozone destroyed:
PV = nRT
Rearranging the equation, we get:
V = nRT / P
Where:
V = volume of ozone destroyed
n = moles of O3 destroyed
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
P = pressure in atm
Plugging in the values:
V = (1.676 mol) x (0.0821 L·atm/(mol·K)) x (235 K) / (25.0 mmHg)
V = 4.032 L
Therefore, the total volume of ozone destroyed when all of the chlorine from 15.5g of CF3Cl goes through ten cycles of the reactions is 4.032 L.
To answer this question, we need to determine the number of moles of CF3Cl and calculate the number of moles of chlorine produced from it. We can then use stoichiometry to determine the number of cycles needed to react all the chlorine produced with ozone.
1. Calculate the number of moles of CF3Cl:
First, we need to determine the molar mass of CF3Cl (Chlorotrifluoromethane) by adding the atomic masses of its constituent elements.
C = 12.01 g/mol
F = 19.00 g/mol (3 fluorine atoms)
Cl = 35.45 g/mol
Molar mass of CF3Cl = 12.01 + 19.00 + 19.00 + 19.00 + 35.45 = 104.46 g/mol
Now, we can calculate the number of moles of CF3Cl:
moles = mass / molar mass = 15.5 g / 104.46 g/mol = 0.148 moles
2. Calculate the number of moles of chlorine (Cl) produced from CF3Cl:
From the balanced equations, we can see that for every mole of CF3Cl, one mole of chlorine is produced.
Therefore, the number of moles of chlorine produced = 0.148 moles
3. Determine the number of cycles needed to react all the chlorine produced with ozone:
From the given set of equations, it takes 1 cycle to react one mole of chlorine with ozone.
So, to react all the chlorine produced, we would need 0.148 moles x 10 cycles = 1.48 moles x cycles.
4. Calculate the volume of ozone destroyed:
First, convert the pressure from mmHg to atm (using 1 atm = 760 mmHg):
25.0 mmHg / 760 mmHg/atm = 0.0329 atm
Now, we can use the Ideal Gas Law equation to determine the volume of ozone destroyed:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L.atm/mol.K), and T is the temperature.
Rearranging the equation to solve for V:
V = (nRT) / P
V = (1.48 mol x 0.0821 L.atm/mol.K x 235 K) / 0.0329 atm
V = 10.6 L
Therefore, the total volume of ozone destroyed when all of the chlorine from 15.5 g of CF3Cl goes through ten cycles of the given reactions would be 10.6 liters.