Question

Metal M reacts with HCl to produce hydrogen gas and a solution of MCl3. If .724 g of M requires 27.62 mL of 1.129 M HCl what is the molar mass of M?
Given reaction: 2M(s)+6HCl(aq) --> 3H2(g)+2MCl3
I know you have to convert 27.62 mL to L, which is .02762 L; and I also found the mol of HCl which is 31.183, but I don't know where to go from there.

Answer 1

Keyword is molar mass, whose units are grams/mole.

You know the first step which is to find moles of HCl, but instead of 31.183 mol HCl, I got (1.129 M HCL * .02762 L) = .03118298 mol HCl. I think you forgot to convert ml->L here. Then, from here we know the amount of moles of HCl, but looking at the equation, it takes 6 moles of HCl to react with 2 moles of metal M so there's some stoichiometry involved. And you want to know the amount of moles of metal M in order to find it's molar mass.

From here, you would multiply .03118298 mol HCl with the stoichiometric ratio (2 mol M for every 6 mol HCl), and the moles HCl would then cancel out and you would get mol M.
So... .03118298 mol HCl * (2 mol M/6 mol HCl) = 0.0103943267 mol of metal M.

Okay so now, you have grams of metal M, and also moles of metal M. You want it's molar mass, units being g/mol. Just divide grams by moles to get your answer.

This is what I got.
(0.724 g M/0.0103943267 mol M) = 69.65338143 g/mol M. Rounded to 3 sig figs would be 69.7 g/mol metal M.

Please check my answer and work, and see if you follow through with the same answer.