Mc. Q9 (1 Viewer)

[Komaticom]

I couldn't do this one.
How does knowing the mass of NO and O3 help in finding mass (hence volume) of NO2?

 

[dingding]

you work out the moles of the reactants, then find which is the limiting reactant, then you know how many moles of product and therefore volume.

 

[xeriphic]

this case you have to do the above two times for NO2 and O3, you would notice O3 with the lower value hence the answer, limiting reagent of the reaction

 

[KungPow]

Ah crap that doesn't sound like what I did. What was the final answer then?

 

[Komaticom]

I'm lost. Working out would be nice.

 

[Steven12]

final answer was A

I put D cos i didnt realise the limiting reagen. i am doomed

 

[scruffy012]

i think i put C but i dunno, chem has eft my mind

scruffy

 

[KungPow]

Well I had A and what I assumed was:

The mass of the two reactants would be the same as the two products.
Thus NO2 + O2 = 0.66 + 0.72
Then I just assumed that the mass of NO2 was half of that.
Then you can work out number of moles, since you have mass and molar mass (work it out)
Then with number of moles you can work out volume (by multiplying with 22.71)

Yeah I know crazy way of thinking but hey, it got me an answer that was one of the choices, which is more than i can say for the other question!

 

[speedy3020]

Well i dont know what limiting reagent you're talking about steven12 but i ended up with something like 0.4995 so D at 0.50L

 

[sub]

A!!! YES!! right again...suk that chem...sub is moving up in the world! :uhhuh:
but i screwed the rest up

 

[Timmy-V]

Well I had A and what I assumed was:

The mass of the two reactants would be the same as the two products.
Thus NO2 + O2 = 0.66 + 0.72
Then I just assumed that the mass of NO2 was half of that.
Then you can work out number of moles, since you have mass and molar mass (work it out)
Then with number of moles you can work out volume (by multiplying with 22.71)

Yeah I know crazy way of thinking but hey, it got me an answer that was one of the choices, which is more than i can say for the other question!

Yeah that is what I did... It seems pretty much right too me.

 

[tennille]

I put A. But I guessed. How lucky...

 

[Komaticom]

Did anybody here get this question right without guessing any single step in the calculation process?!

 

[Brad]

I Just thought that since O2 gave the lower value or someshit, that it was the limiting reagent and therefore the reaction could not continue without any moles of O2 left therefore the maximum amount of The product is the same number of moles as there is of O2

n=0.72/48
n=0.015
0.015=m/(whatever the molar mass of NO2 is)
Which gave A

 

[Komaticom]

Thanks Brad.
Can't rep ya, argh. Have to spread some around first. *grumble grumble*

 

[sub]

I Just thought that since O2 gave the lower value or someshit, that it was the limiting reagent and therefore the reaction could not continue without any moles of O2 left therefore the maximum amount of The product is the same number of moles as there is of O2

n=0.72/48
n=0.015
0.015=m/(whatever the molar mass of NO2 is)
Which gave A

uhh, dude its O3...

 

[lucyinthehole]

i did some crazy calculations that even i can't remember, that gave me 0.68. i divided that by 2 to get 0.34. i was more striving towards one of the answers, as opposed to actually knowing how to do it

 

[bloodysunday]

i suck at all of the calculation questions... hmm thats bad huh lol

 

[mervvyn]

I'm with brad.... O3 is limiting reagent, work out moles and go from there.

 

[lukebennett]

you just had to realise that you needed ratio of mole 1:1, which meant that you were limited by the smallest number of moles. if different number of moles react the ratio in the reaction would be wrong