I have found mols of HCl and I have found mols of the unknown Metal (M). What are my next steps?

Question

I have found the mols of the HCl solution and I have found the mols of the unknown M. I divided by the mass to get 53.89g/mol. What I did next was find the mols of NaOH by using the formula M=mols/L. What I calculated was 0.00224mols of NaOH. I then took my 53.89g/mol and divided that by the 0.00224mols of NaOH. From that I got a ridiculous number that does not even come close to the numbers on the periodic table. Where did I go wrong in my steps? 

 

Balanced chemical equation: 2M(s) + 6HCl(aq) ==> 2MCl3 + 3H2(g)

 

Question: 0.0900 g of metal is reacted with 50.00 mL of 0.100 M HCl. Once the reaction is complete, the reaction mixture is titrated with 0.1375 M NaOH. It requires 16.35 mL of NaOH to reach the endpoint. What is the molecular weight of the metal that was reacted?

Answer 1

Not really following your explanation.  How did you get 53.89 g/mol?  That would be the molecular weight.  How did you find the moles of unknown?  If you have the moles of unknown, you should divide the mass of the unknown by the moles to get the molecular weight.  Your calculation of moles of NaOH is correct but why you divided 53.89 g/mole by that number is not clear.  That makes no sense.

 

This problem is one of back-titration.  The moles of NaOH used is the same as the moles of HCl remaining after the reaction (there is excess HCl).  From the moles of HCl initially added (0.05 L x 0.100 mol/L) and the moles left over, you can calculate the moles of HCl used.  From that, you can determine moles of M present (using the stoichiometry of the balanced equation).  Once you have moles of M, you can use 0.09 g M/moles M = molecular weight of M.

 

Let me know if you need more help.