What is the percent by mass of the magnesium hydroxide?

Let us first work out how many moles of acid were neutralized by the sample.

Beginning:

#0.1 l\times{0.2050" mol HCl"}/l=0.02050# mol #"HCl"#

After the sample was added:

We are given the amount of #"NaOH"# that would neutralize the remaining #"HCl"#:

#0.01985 l\times{0.1020 "mol NaOH"}/l=0.0020247# mol #"NaOH"#

Now each me of the added #"NaOH"# neutralized one mole of leftover #"HCl"# from the balanced equation

#"NaOH"+"HCl"->"NaCl"+"H"_2"O"#

so we have:

#0.0020247# mol #"NaOH"\times{1" mol HCl"}/{1" mol NaOH"}=0.0020247# mol #"HCl"# left over

So we started with 0.02050 mol #"HCl"# and we were left with 0.0020247 mol #"HCl"# to after the sample was added. The difference, rounded to the smaller number of decimal places, is:

#0.02050-0.00202=0.01848 "mol HCl"# that the sample neutralized.

Now we can work out how mych magnesium hydroxide must have neutealuzed that acid. Start with the balanced equation:

#"Mg(OH)"_2+2 "HCl"->"MgCl"_2+2 "H"_2"O"#

So then:

#0.01848" mol HCl"\times{1" mol Mg(OH)"_2}/{2" mol HCl"}=0.009240 "mol Mg(OH)"_2# in the sample

We weighed the sample in grams so calculate the grams of magnesium hydroxide using the molecular or formula weight #=58.3197" g/mol"#. So

#0.009240 "mol"\times58.3197" g/mol"=0.5389 g#.

This is how much magnesium hydroxide is in the #0.5895 "g"#'sample. Then

#0.5389 "g"/0.5895"g"=0.9142=91.42%#