A sample of impure magnesium was allowed to react with excess HCL solution. After 5.5 gram of the impure metal was treated with 0.3L of 0.6 M HCL, 0.0125 mole of HCL remained. Assuming the impurities do not react with the acid, what is the mass % of magnesium in the sample?

Answer 1

The Mg metal (the real stuff) reacts with the HCl as follows:
2HCl + Mg ==> MgCl2 + H2
moles HCl initially = M x L = 0.6 M x 0.3 L = 0.18 mols.
moles HCl remaining after reaction = 0.0125
moles HCl used in the reaction = 0.18 - 0.0125 = 0.1675
Convert 0.1675 mols HCl used to Mg used as follows:
0.1675 mols HCl x (1 mol Mg/2 mol HCl) = 0.08375 mols Mg used.
Convert that to grams. g = mols x atomic mass Mg = 2.03 g Mg
% Mg in the original sample = (mass Mg/mass sample)*100 = (2.03/5.5)* 100 = ? about 37%. Note: I have used more significant figures than allowed because the problem makes no effort to distinguish between 0.6 M and 0.3 L vs. 0.600 M and 0.300 L.