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

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 [tex]Cl_2\text{ and }F_2[/tex]

Using ideal gas equation:

[tex]n_{Cl_2}=\frac{P_{Cl_2}V}{RT}[/tex]

where,

[tex]P_{Cl_2}[/tex] = partial pressure of [tex]Cl_2[/tex] = 337 mmHg = 0.443 atm

R = gas constant = 0.0821 L.mmHg/mol.K

T = temperature = 298 K

V = volume = 1.87 L

[tex]n_{Cl_2}=\frac{(0.443atm)\times (1.87L)}{(0.0821L.atm/mol.K)\times (298K)}[/tex]

[tex]n_{Cl_2}=0.0338mol[/tex]

and,

[tex]n_{F_2}=\frac{P_{F_2}V}{RT}[/tex]

where,

[tex]P_{F_2}[/tex] = partial pressure of [tex]F_2[/tex] = 820 mmHg = 1.08 atm

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 298 K

V = volume = 1.87 L

[tex]n_{F_2}=\frac{(1.08atm)\times (1.87L)}{(0.0821L.atm/mol.K)\times (298K)}[/tex]

[tex]n_{F_2}=0.0825mol[/tex]

Now we have to calculate the limiting and excess reagent,

The balanced chemical reaction is,

[tex]Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)[/tex]

From the balanced reaction we conclude that

As, 3 mole of [tex]F_2[/tex] react with 1 mole of [tex]Cl_2[/tex]

So, 0.0825 moles of [tex]F_2[/tex] react with [tex]\frac{0.0825}{3}=0.0275[/tex] moles of [tex]Cl_2[/tex]

From this we conclude that, [tex]Cl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]F_2[/tex] is a limiting reagent and it limits the formation of product.

Renaming moles of [tex]Cl_2[/tex] = 0.0338 - 0.0275 = 0.00630 mol

Now we have to calculate the partial pressure of excess reactant after the reaction.

[tex]P_{Cl_2}=\frac{n_{Cl_2}RT}{V}[/tex]

[tex]P_{Cl_2}=\frac{(0.00630mol)\times (0.0821L.atm/mol.K)\times (298K)}{1.87L}[/tex]

[tex]P_{Cl_2}=0.0824atm=0.0824\times 760mmHg=62.6mmHg[/tex]

Thus, the partial pressure of excess reactant after the reaction is, 62.6 mmHg