Chemistry, 02.03.2020 23:03, angeleyes4u610p6np54

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

Answers: 1

Show answers

Answers

answered: zoeyruckel

This is an incomplete question, here is the complete question.

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)$

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 $Cl_2\text{ and }F_2$

Using ideal gas equation:

$n_{Cl_2}=\frac{P_{Cl_2}V}{RT}$

where,

$P_{Cl_2}$ = partial pressure of Cl_2 = 337 mmHg = 0.443 atm

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

T = temperature = 298 K

V = volume = 1.87 L

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

$n_{Cl_2}=0.0338mol$

and,

$n_{F_2}=\frac{P_{F_2}V}{RT}$

where,

$P_{F_2}$ = partial pressure of F_2 = 820 mmHg = 1.08 atm

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

T = temperature = 298 K

V = volume = 1.87 L

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

$n_{F_2}=0.0825mol$

Now we have to calculate the limiting and excess reagent,

The balanced chemical reaction is,

$Cl_2(g)+3F_2(g)\rightarrow 2ClF_3(g)(s)$

From the balanced reaction we conclude that

As, 3 mole of F_2 react with 1 mole of Cl_2

So, 0.0825 moles of F_2 react with $\frac{0.0825}{3}=0.0275$ moles of Cl_2

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

Renaming moles of Cl_2 = 0.0338 - 0.0275 = 0.00630 mol

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

$P_{Cl_2}=\frac{n_{Cl_2}RT}{V}$

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

$P_{Cl_2}=0.0824atm=0.0824\times 760mmHg=62.6mmHg$

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

answered: Guest

The nucleus inside an atom consists of positively-charged protons and neutrally-charged neutrons. a proton is a subatomic particle with a positive electric charge. allprotons are identical.

answered: Guest

the tides are influenced by both the moon and the sun, it's easy to see that when the sun lines up with the moon and the earth, as during a new moon or full moon, the tidal effect is increased. these are known as spring tides, named not for the season, but for the fact that the water "springs" higher than normal. on the other hand, if the sun and the moon are 90 degrees apart in relation to an observer on earth as during the first quarter moon or third quarter moon, then high tides are not as high as they normally would be. this is because despite its greater distance, the sun's mass allows it to exert enough gravitational force on the oceans that it can negate some of the effects of the moon's pull. this phenomenon of lower high tides is called a neap tide. if you need more information on this topic visit moonconnection.com.