Question

Carbon mono oxide (CO) and oxygen O2 react according to
2CO(g) + O2(g)
→ 2CO2(g)
Assuming that the reaction takes place and goes to completion, after the
valve is opened in the apparatus represented in the accompanying figure.
3 litre
Also assume that the temperature is fixed at 300 K. (Take R = 0.08 atm
at 16 atm
L/mole K)
(A) Partial Pressure of O2 = 6 atm.
(B) Number of moles of CO2 formed = 2
) Number of moles of Oz left = 1
(D) Partial Pressure of O2 - 3 atm.
at 48 am​

Answer 1

Answer:

Partial pressure of O2 = 5.33 atm

Explanation:

Calculate the total number of moles first, with PV= nRT

Given ,

P = 16 atm

R = 0.08

V= 3 L

Number of moles of O2 reacted be X and moles of CO will be 2x ( from the reaction)

Now,

$PV= n} total R Tor, 16 * 3 = ntotal * 0.0821 * 300On solving , we get ntotal = 1.95 moles Now, X+ 2X = 1.95X= 0.65 Thus Moles of O2 = 0.65 \\Moles of CO = 1.3 moles of CO2 formed = 1.3 moles[tex]Pressure exerted by O2 = \frac{nRT}{V} \\= \frac{0.65*0.0821*300}{3}= 5.33 atm The answer A seems correct.$