It should be lithium.

Another source that I considered trusted down the drain

I have seen this question is several places. In one of them it was printed in two versions - one called for addition of Me

_{2}O for neutralization, other for addition on MeOH, both were claimed to yield the same result. Turns out MeOH version is wrong, but I have not checked it before posting - I have solved Me

_{2}O version only.

For Me

_{2}O and starting from very similar equations you both did it is possible to show that mass of HX must equal

[tex]m_{HX} = \frac {100} {9} (R_{Me}-1)[/tex]

That's very similar to the latest equation listed by Dan. 100 is a mass of water (not of the solution!) in the HX solution, 9 is half of the molar mass of water, R

_{Me} is molar mass of metal. For lithium that yields around 66.6 gram of HX, and concentration of

[tex]\frac{66.6}{100+66.6}100\% = 40\%[/tex]

and over 70% for sodium - so it can be safely assumed it must be Li, as 70% solutions won't be stable at STP.

Sadly, it doesn't work this way for MeOH, as you have both shown.

Edited the original wording of the problem.