[Astrokel]

hello,

i am stuck in this question as attached and i have no idea how to start. can anyone start me off with some hints or explanations?

my attempt: 2C6H5N + 33/2 O2 ---> 5H2O + 12CO2 + 2NO2 (+excess oxygen)
calcium chloride will remove the water, while KOH remove acidic gases nitrogen dioxide and carbon dioxide. So at point Q will left with excess oxygen while at P all the 4 products will be presence. Then isn't Mp/Mq  > 1 since mass at Mp is definitely larger due to all 4 products while at Q only excess oxygen, so answer will be D, but it is not.

thank you

[Borek]

Think it over, you are on the wrong track.

They ask not about gases present, but about gases adsorbed. Just look at products and think where they will end.

Also note that they want you to assume that NO₂ was not adsorbed. Honestly, I have no idea if it will react with KOH or not. For sure it not precursor of nitrate nor nitrite, but that's chemistry, strange things may happen.

[Astrokel]

thanks for the response

Did i miss out something on the assumption of NO₂ was not adsorbed, or is it because the reaction is unsure of?

looking at the products, the increase in vessel of Q will be due to CO₂ and P will be H₂O

Mr of CO2 = 44 , H2O = 18

Mp/Mq = (18x5)/(44x12) = 0.17

ye!!!!!!!!! thanks Borek for the guiding, thats the right answer

[Borek]

Did i miss out something on the assumption of NO₂ was not adsorbed, or is it because the reaction is unsure of?

Results show that you have to assume it is not adsorbed (yes, I have solved the question before answering to check). Also, I don't remember reading or learning about NO₂ reacting with bases; but first - I don't know everything, second - this is chemistry, doesn't matter what you think, what matters is what happens in the glass.

[Astrokel]

thanks heaps!!