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

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... =/

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:
C_n H_2n + H_2 \rightarrow C_n H_{2n+2} but from now on I don't know what to do.

Thank you, anyway! :)

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?

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

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.

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.

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

C_nH_{2n} + H_2 \rightarrow C_nH_{2n+2}

C_nHN{2n} = 12n+2n=14n
C_nH_{2n+2} = 12n + 2n +2=14n+2

14n...................22.4 L
x.......................2.24 L (there is mistake in the problem)
x=1.4 n

1.4n(14n+2)=14n*3
19.6n + 2.8 = 42
n=\frac{42-2.8}{19.6}=2
the forumula is: C_2H_4

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