# 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.

May 15, 2016

The sample is 91.42% magnesium hydroxide.

#### Explanation:

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

Beginning:

$0.1 l \setminus \times \frac{0.2050 \text{ mol HCl}}{l} = 0.02050$ mol $\text{HCl}$

We are given the amount of $\text{NaOH}$ that would neutralize the remaining $\text{HCl}$:

$0.01985 l \setminus \times \frac{0.1020 \text{mol NaOH}}{l} = 0.0020247$ mol $\text{NaOH}$

Now each me of the added $\text{NaOH}$ neutralized one mole of leftover $\text{HCl}$ from the balanced equation

$\text{NaOH"+"HCl"->"NaCl"+"H"_2"O}$

so we have:

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

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

$0.02050 - 0.00202 = 0.01848 \text{mol HCl}$ that the sample neutralized.

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

$\text{Mg(OH)"_2+2 "HCl"->"MgCl"_2+2 "H"_2"O}$

So then:

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

We weighed the sample in grams so calculate the grams of magnesium hydroxide using the molecular or formula weight $= 58.3197 \text{ g/mol}$. So

$0.009240 \text{mol"\times58.3197" g/mol} = 0.5389 g$.

This is how much magnesium hydroxide is in the $0.5895 \text{g}$'sample. Then

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