This is an incomplete question, here is the complete question.
Chlorine gas reacts with fluorine gas to form chlorine trifluoride.
![Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)](https://tex.z-dn.net/?f=Cl_2%28g%29%2B3F_2%28g%29%5Crightarrow%202ClF_3%28g%29%28s%29)
A 1.87 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 820 mmHg . What is the partial pressure of excess reactant after the reaction occurs as completely as possible?
Answer : The partial pressure of excess reactant after the reaction is, 62.6 mmHg
Explanation :
First we have to calculate the moles of ![Cl_2\text{ and }F_2](https://tex.z-dn.net/?f=Cl_2%5Ctext%7B%20and%20%7DF_2)
Using ideal gas equation:
![n_{Cl_2}=\frac{P_{Cl_2}V}{RT}](https://tex.z-dn.net/?f=n_%7BCl_2%7D%3D%5Cfrac%7BP_%7BCl_2%7DV%7D%7BRT%7D)
where,
= partial pressure of
= 337 mmHg = 0.443 atm
R = gas constant = 0.0821 L.mmHg/mol.K
T = temperature = 298 K
V = volume = 1.87 L
![n_{Cl_2}=\frac{(0.443atm)\times (1.87L)}{(0.0821L.atm/mol.K)\times (298K)}](https://tex.z-dn.net/?f=n_%7BCl_2%7D%3D%5Cfrac%7B%280.443atm%29%5Ctimes%20%281.87L%29%7D%7B%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28298K%29%7D)
![n_{Cl_2}=0.0338mol](https://tex.z-dn.net/?f=n_%7BCl_2%7D%3D0.0338mol)
and,
![n_{F_2}=\frac{P_{F_2}V}{RT}](https://tex.z-dn.net/?f=n_%7BF_2%7D%3D%5Cfrac%7BP_%7BF_2%7DV%7D%7BRT%7D)
where,
= partial pressure of
= 820 mmHg = 1.08 atm
R = gas constant = 0.0821 L.atm/mol.K
T = temperature = 298 K
V = volume = 1.87 L
![n_{F_2}=\frac{(1.08atm)\times (1.87L)}{(0.0821L.atm/mol.K)\times (298K)}](https://tex.z-dn.net/?f=n_%7BF_2%7D%3D%5Cfrac%7B%281.08atm%29%5Ctimes%20%281.87L%29%7D%7B%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28298K%29%7D)
![n_{F_2}=0.0825mol](https://tex.z-dn.net/?f=n_%7BF_2%7D%3D0.0825mol)
Now we have to calculate the limiting and excess reagent,
The balanced chemical reaction is,
![Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)](https://tex.z-dn.net/?f=Cl_2%28g%29%2B3F_2%28g%29%5Crightarrow%202ClF_3%28g%29%28s%29)
From the balanced reaction we conclude that
As, 3 mole of
react with 1 mole of ![Cl_2](https://tex.z-dn.net/?f=Cl_2)
So, 0.0825 moles of
react with
moles of ![Cl_2](https://tex.z-dn.net/?f=Cl_2)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Renaming moles of
= 0.0338 - 0.0275 = 0.00630 mol
Now we have to calculate the partial pressure of excess reactant after the reaction.
![P_{Cl_2}=\frac{n_{Cl_2}RT}{V}](https://tex.z-dn.net/?f=P_%7BCl_2%7D%3D%5Cfrac%7Bn_%7BCl_2%7DRT%7D%7BV%7D)
![P_{Cl_2}=\frac{(0.00630mol)\times (0.0821L.atm/mol.K)\times (298K)}{1.87L}](https://tex.z-dn.net/?f=P_%7BCl_2%7D%3D%5Cfrac%7B%280.00630mol%29%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28298K%29%7D%7B1.87L%7D)
![P_{Cl_2}=0.0824atm=0.0824\times 760mmHg=62.6mmHg](https://tex.z-dn.net/?f=P_%7BCl_2%7D%3D0.0824atm%3D0.0824%5Ctimes%20760mmHg%3D62.6mmHg)
Thus, the partial pressure of excess reactant after the reaction is, 62.6 mmHg