I though the problem is finished...

so there is 0.240328mol of H2 liberate by PV=nRT

(Taking its Ga and Al-anyway it seems to be okay because it's group III: so reaction is

M + 3H20---> 1.5H2 + M(OH)3 or M + 3HCl---> 1.5H2 + MCl3 (just take acid as HCl then its easier to calc...)

If its totally Al then there will be 4.16g of the alloy, if its Ga then 11.05g of alloy. So it makes cents as 5 is between. (though it doesn't make dollars haha!)

Take there is X mol of Al, Ymol of Ga.

Then 26.98X+69.72Y=5.00,

1.5(X +Y) = 0.240328

solving stimutaneous equations X=0.14437, Y=0.015847

So there's 0.14437mol of Al (3.8951g) and 0.015847mol of Ga (1.1049g)

I hope this is correct... Whee this is so fun! Will definitely look out for the next one!

EDIT: Which seems like it corresponds quite nicely to the 30-70 ratio stated in the paper.