If it takes 50 mL of 0.05M HCL to neutralize 345 mL of Mg(OH)2 solution, what is the concentration of the Mg(OH)2 solution?

First, write the unbalanced equation:
Mg (OH)_2 + H Cl = Mg Cl_2 + H_2 O.
It is simple to balance it:
Mg (OH)_2 + 2H Cl = Mg Cl_2 + 2H_2 O
(give the coefficient 2 to H Cl to balance 2 Cl at the right side, then give the coefficient 2 to H_2 O to balance O, and we are done).
Thus one mole of Mg (OH)_2 requires two moles of H Cl to be neutralized.
Next, there are 0.05 (mol)/L * 50*10^(-3) L = 2.5*10^-3 mol  of  H Cl  in  H Cl solution, therefore there are twice less moles of Mg (OH)_2, 1.25*10^-3 mol. This gives the molarity (molar concentration) of Mg (OH)_2 to be equal to  (1.25*10^-3 mol)/(345 mL) = (1.25 mol)/(345 L) approx 0.0036 M.  This is the answer.