[Rutherford]

An unknown substance reacts with water, H₂ is released with only one another product- a soluble base of a metal. When you in 50cm³ of water add 10g of the unknown substance, the mass share from the product is 28.17%. What is the unknown substance?

By my thinking, the substance is a metal or a oxide or hydride of a alcaline or earth-alcaline metal.
But I don't know how to determine it. How to use the info that is given? I don't know how to write the reaction because I don't know the molar ratio of the unknown substance and water that reacts.

[Borek]

Partially trial and error. Assume it was a metal. Check if you can use the data given to find equivalent mass of the metal. Does it fit any metal?

I must say wording of the question is not clear to me, but I guess it is just a matter of your translation.

[Rutherford]

Just a metal won't work. What isn't clear?

[Borek]

Just a metal won't work.

Yes it will. Try. If you have no idea how, try assuming it is a monovalent metal.

What isn't clear?

This part:

When you in 50cm3 of water add 10g of the unknown substance, the mass share from the product is 28.17%.

[Rutherford]

1st the translation: I think that it is so said in my mother language, only "the mass share from the product is 28.17%".If I made a mistake I am sorry. I suppose that they mean the base that is made because H₂ vaporizes.

I tried with a monovalent metal:
2M+2H₂O-->2MOH+H₂
M is the mass of the metal.
n(H₂)=(10/M)/2 m(H₂) is 10/M.
m(MOH)=(10/M)*(M+17)=(10M+170)/M
Now:
28.17%:100%=m(MOH):[(50+10)-m(H₂)], 50 is the mass of water (d=1g/cm³).
M=24.6g doesn't fit to any metal. The nearest one is Mg but Mg is divalent.

Then I tried with the oxides, but they don't produce H₂ when solubled in water.

Only hydrids left. I tried hydrid of a monovalent metal and I got that M=23g(Na). So the substance would be NaH. The problem is I don't have the answer, so if someone could confirm that I done this well.

[Borek]

Have you tried with divalent metals?

Can you post the phrase in original?

[DrCMS]

I agree with NaH being the correct answer -  NaH gives 28.17% or 27.78% depending if you only count the solution rather than all the products.  So I agree that that the phrase

"the mass share from the product is 28.17%"

is open to interpretation.

[Rutherford]

Ok, thanks.

[Olympiad_Tutor]

I smell an olympiad question here. 

Hydride was a good idea.
Nothing else that could give a soluble M(OH)n seems to match the numbers.

what's the original language of the problem?

[Rutherford]

My state doesn't compete in the IChO. This is only a preparation task for the state competition.

[Rutherford]

Have you tried with divalent metals?

Can you post the phrase in original?

Oh, didn't see this. Yes I tried with divalent metals and I got that M=25.44g.