Question

What is the percent by mass of the magnesium hydroxide?

A 0.5895 g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid needs 19.85 mL of 0.1020 M NaOH for neutralization.

Answer 1

The sample is 91.42% magnesium hydroxide.

Explanation:

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%`