FWS asked in Science & MathematicsChemistry · 1 month ago

# a. Calculate the number of moles of substance that remains.  b. What are the partial pressures of the gases after the valve is opened?

Assuming that the reaction takes place in the apparatus shown above at 300 K and goes to completion after the valve is opened, determine the following:

Relevance
• 1 month ago

Initial moles:

PV = n RT

For CO:

0.500 atm (2.00 L) = n (0.08206 Latm/molK)(300K)

n = 0.0406 mol CO

For O2:

1.00 atm (1.00 L) = n (0.08206 Latm/molK)(300K)

n = 0.0406 mol O2

CO is the limiting reactant, so,

0.0406 mol CO X (2 mol CO2 / 2 mol CO) = 0.0406 mol CO2 is formed

Moles O2 consumed:

0.0406 mol CO X (1 mol O2 /2 mol CO) = 0.0203 mol O2 consumed, so 0.0203 mol O2 remains

So, after the reaction:

For CO2:

P (3.00 L) = 0.0406 mol (0.08206 Latm/molK) (300 K)

P = 0.333 atm CO2

For O2:

P (3.00 L) = 0.0203 mol (0.08206 Latm/molK)(300 K)

P = 0.167 atm O2

• Dr W
Lv 7
1 month ago

let's try it a different way than everyone else here is doing..... we know that for gases with the same volume and temperature, we can read coefficients of the balanced equation as PRESSURE ratios as well as mole ratios..

********

once we open the valve,

.. P CO = 0.500atm * (2L / 3L) = 1/3 atm

.. P O2 = 1.00atm * (1L / 3L) = 1/3 atm

from the balanced equation

.. 1/3 atm CO * (2 atm CO2 / 2 atm CO) = 1/3atm CO2

.. 1/3 atm O2 * (2 atm CO2 / /1 atm O2 = 2/3 atm CO2

therefore

.. O2 is the LR and 1/3 atm CO2 is formed

after rxn

.. atm CO remaining = 0

.. atm O2 consumed = 1/3atm CO2 * (1 mol O2 / 2 mole CO2) = 1/6 atm O2

.. atm O2 remaining = 1/3 atm - 1/6atm = 1/6 atm

.. atm CO2 formed = 1/3 atm

so let's see.. that give us "b".. let's do a.

.. n O2 = PV/(RT) = 1/6atm * 3L / (0.08206Latm/molK  * 300K) = 0.0203

.. n CO2 = 1/3atm * 3L / (0.08206Latm/molK * 300K) = 0.0406

• 1 month ago

a.

(2.00 L) x (0.500 atm) = 1.00 L atm CO

(1.00 L) x (1.00 atm) = 1.00 L atm O2

1.00 liter-atmosphere of CO would react completely with 1.00 x (1/2) = 0.50 liter-atmosphere of O2, but there is more O2 present than that, so O2 is in excess and CO is the limiting reactant.

(1.00 L atm O2 originally) - (0.50 L atm O2 reacted) = 0.50 L atm O2 remaining

n = PV / RT = (0.50 L atm) / ((0.082057366 L atm/K mol) x (300 K)) =

0.0203 mol O2 remaining

b.

Since it is the limiting reactant, the partial pressure of CO is zero after the reaction is complete.

(0.50 L atm O2) / (2.00 L + 1.00 L) = 0.17 atm of leftover O2

(1.00 L atm CO) x (2 mol CO2 / 2 mol CO) / (2.00 L + 1.00 L) = 0.33 atm of CO2

• david
Lv 7
1 month ago

CO --  n = PV/(RT)  = 0.500X2.00/[0.0821X573]  =  0.02126 mol CO << initial amunt befr reaction

O2 -- n = (1.00X1.00)/[0.0821X573]  =  0.02126 mol O2 >> initial

a.   1/2 of the O2 is used  --  leaving  0.01063 mol O2  <<< answer

and creating 0.02126 mol CO2  << answer

b.   P = nRT/V

P(O2)  =  0.01063X0.0821X573/3.0  =  0.1667 atm

P(CO)  =  0.02126X0.0821X573/3.0  =  0.3333 atm

...  round both for sig figs.