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.

1 Answer
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:

0.1l×0.2050 mol HCll=0.02050 mol HCl

After the sample was added:

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

0.01985l×0.1020mol NaOHl=0.0020247 mol NaOH

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

NaOH+HClNaCl+H2O

so we have:

0.0020247 mol NaOH×1 mol HCl1 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:

0.020500.00202=0.01848mol 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+2HClMgCl2+2H2O

So then:

0.01848 mol HCl×1 mol Mg(OH)22 mol HCl=0.009240mol 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

0.009240mol×58.3197 g/mol=0.5389g.

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

0.5389g0.5895g=0.9142=91.42%