Stoichiometry problem & percent by mass?

Question

It takes 37.25 mL of 0.2815 M phosphoric acid (H3PO4) to react completely with a 1.364 g sample, which has some calcium magnesium hydroxide and other inert ingredients. Assuming that the H3PO4 and Mg(OH)2 react according to the reaction below, calculate the grams of magnesium hydroxide in the sample, and the percent by mass of magnesium hydroxide in the sample.

balanced equation:

3 Mg(OH)2 (s) + 2 H3PO4 (aq) -----> 6 H2O (l) + Mg3(PO4)2 (aq)

The answer (for the percent by mass) is 67.27%, I just need help getting there. I can do most stoichiometry problems, but this one is a bit confusing to me (with the sample)..I could just use a push in the right direction.

Any help would be much appreciated. :)

Answer 1

Always start with what you know and in any reaction it is the moles that are important.

so Moles H3PO4 = concentration * vol (in Litres)

= 0.2815 * 0.03725

=0.01049

From the reaction 2 moles of H3PO4 reacts with 3 Moles of Mg(OH)2

So the number of moles Mg(OH)2 reacted =0.01049*3/2

= 0.01573

So mass Mg(OH)2 = moles * molecular weight

=  0.01573 * (24.31 + (1.008 + 16.00)*2)

= 0.9174 g

% mass in Sample = 0.9174/1.364*100

=67.26%