The picture below shows two bulbs connected by a stopcock. The 6.00-L bulb conta

Question

The picture below shows two bulbs connected by a stopcock. The 6.00-L bulb contains nitric oxide at a pressure of 0.400 atm, and the 1.50-L bulb contains oxygen at a pressure of 2.50 atm. After the stopcock is opened, the gases mix and react: 2NO(g) + O_2(g) right arrow 2NO_2(g) Determine which gases remain after the reaction goes to completion and calculate their partial pressures. The temperature at the beginning and the end of the experiment is 22 degree C. Which gases are present at the end of the experiment What are the partial pressures of the gases If the gas was consumed completely, put 0 for the answer.

Explanation / Answer

we know that

PV = nRT

so

For 6 L bulb , it contains N0

0.4 x 6 = n x 0.0821 x 295

n = 0.0991

so

0.0991 moles of NO is present

now

For 1.5 L bulb with 02

2.5 x 1.5 = n x 0.0821 x 295

n = 0.155

now

the reaction is

2 N0 + 02 --> 2 N02

we can see that

moles of 02 reacted = 0.5 x moles of N0

moles of 02 reacted = 0.5 x 0.0991

moles of 02 reacted = 0.04955

but

we have 0.155 moles of 02

so

02 is remaining , also

moles of 02 remaining = 0.155 - 0.04955 = 0.10545

also

moles of N02 formed = moles of N0 reacted = 0.0991

so

at the end of expermient

02 and N02 are present

now

final volume = 6 + 1.5 = 7.5 L

for oxygen

PV = nRT

P x 7.5 = 0.10545 x 0.0821 x 295

P = 0.34 atm

for N02

PV = nRT

P x 7.5 = 0.0991 x 0.0821 x 295

P = 0.32

so

finally

P NO = 0

P 02 = 0.34

P N02 = 0.32

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