The number of moles of Li would be 0.00500 mol and the mass would be 0.0347 g.
There are two reactions taking place.
#2Li + 2H_2O -> 2LiOH + H_2# when the lithium was placed into the water and...
#H^+ + OH^-##-> H_2O# when the acid was added to the resulting solution.
The #H^+# and #OH^-# react in a 1:1 ratio. This tells us that the number of moles of #H^+# used will be equal to the number of #OH^-# moles in solution. Likewise, 2 moles of lithium produces 2 moles of #OH^-#. This is also a 1:1 ratio. As a result, we can say that for every mole of #H^+# used from the acid, one mole of lithium must have been added to the water at the start of the reaction.
Now to calculate.
#1mol//L xx 0.00500 L = 0.00500 mol H^+# #0.00500 mol H^+ = 0.00500 mol Li#
