A sample of impure magnesium was analyzed by allowing it to react with excess HCl solution:

Question

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

After 1.27 g of the impure metal was treated with 0.100 L of 0.768 M HCl, 0.0125 mol HCl remained. Assuming the impurities do not react, what is the mass % of Mg in the sample?

I converted the 1.27gMg to 0.0522moles Mg. Im just not sure where to go from there. please help me in the right direction.
Thanks!

Answer 1

Nope, you can't directly convert the 1.27 g sample to moles of Mg because it is impure or not 100% Mg.

Since there is some HCl left after the reaction, the impure Mg is the limiting reactant and HCl is in excess.
Note that the Mg used was impure.
To get the number of moles, we use its the relationship with Molarity:
M = n/V
where n = moles and V = liters of solution
We get the moles of HCl:
n, HCl = 0.768 * 0.100 = 0.0768 moles HCl
Since not all of this reacted (as we said above, HCl is in excess), we subtract this from the amount of HCl left:
0.0768 - 0.0125 = 0.0643 moles HCl (amount only reacted with Mg)
Using the stoich ratios from the reaction, for every 2 mol HCl reacted, 1 mol Mg reacted. So,
0.0643 moles HCl * (1 mol Mg / 2 mol HCl) = 0.03215 mol Mg (this is pure)
The molar mass of Mg is 24.3 g/mol. So the mass of pure Mg is therefore,
0.03215 mol * 24.3 g/mol = 0.781245 g pure Mg

Finally, we can get the % pure Mg in the sample:
0.781245 / 1.27 * 100 = 61.5 %

Hope this helps~ :3

Answer 2

You erred in calculating mols Mg.
What you have written is true only if the Mg metal were pure; however, it isn't.
mols HCl begin = M x L = 0.768 x 0.1 = ?
mols HCl remain = 0.0125
the difference is mols HCl used.
Then mols Mg is 1/2 that of HCl.
g Mg is mols x atomic mass Mg
%Mg = (g Mg/mass sample)*100 = ?
I get something like 61% but that isn't exact.