[lemon14]

After the hydrogenation of 22.4 L of an alkene A results 3 g of alkane B. How to find the alkene?

[Unknown008]

Is this the whole problem? I'm not sure I understand...

Okay, it seems that you have 1 mol of an alkene at stp (22.4 L is one mole of gas at stp)

When you hydrogenate this, you get 3 g of alkane?

That's very little... it's less than a mole of carbon while we had exactly one at the start... =/

[lemon14]

Yes, this is the whole problem. I know that it doesn't give much information, but I thought I was missing something.

At first I thought about writing the reaction:
but from now on I don't know what to do.

Thank you, anyway! :)

[Unknown008]

You are sure there is nothing more mentioned, no conditions to be assumed (like pressure, temperature)?

And just in case, I misunderstand, is the 22.4 L the volume of alkene or the volume of hydrogen gas?

[lemon14]

22.4 L is the volume of alkene and the reaction is in stp.

[Unknown008]

Okay... this will mean you have an initial number of moles of alkene of 1.

This will yield 1 mole of alkane, or is there a specific yield involved?

Because 3 g is really small... even 1 mol of alkene doesn't have such a mass.

[lemon14]

No, there isn't a specific yield of alkane involved, only those 3 g.

Maybe the problem is mistaken. Don't bother about it anymore. I'll post the solution if I succeed in find it.

[Unknown008]

Okay, because this question really seems to have something wrong. I'll be waiting :)

[lemon14]

(there is mistake in the problem)





the forumula is:

[Unknown008]

Well, that was the problem, there were initially 0.1 mol of A, reacting with excess H2 gas to form 3 g of alkane.