stoichiometry help, please!?

The balanced chemical equation tells you that 1 molecule of NaOH reacts with 1 molecule of HCl to yield 1 molecule of H2O and 1 molecule of NaCl.

Since 6.022x10^23 molecules = 1 mole, we can substitute "mole" for "molecule" in the above sentence.

1 mole of NaOH = 1 gram molecular weight = 40.0 g/mole
1 mole of HCl = 36.46 g/mol
1 mole H2O = 18.02 g/mol

Example: You started with 4.00 gram NaOH and an excess of HCl.
4.00 g NaOH / 40.0 g/mol =  0.100 mol NaOH.

From the balanced chemical equation you know that 0.100 mole of NaOH will yield 0.100 mole of H2O.

Moles x grams/mole = grams
0.100 mole H2O x 18.02 g/mol = 1.802 gram H2O

Il start with your second question first
Using n=m/mr
n=20/51
n=0.39mol of MgOH.

As the ratio is 1:1, the number of moles is the same so the number of moles of HCl is 0.39.
So 0.39 moles of HCl will neutralize 20g of MgOH

I presume that the experiment is using 20g of so number of moles of water made is 0.39.