 Javan P.

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

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?

By:

Javan P.

I found the mols of HCl from the 50mL and 0.100m HCl which was 0.005mols of HCl. I then took that number and did 0.0005mols and found mols of M from the balanced equation. I then took that number which was 0.00167 mols of M and divided that to get my mass which was 53.89 g. After that is where I got confused.
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10/29/17 J.R. S.

tutor
Look at my previous comment.  The moles of HCl you calculated (0.005) are the initial moles present. But some reacted with M to make MCl3 and H2 gas.  The HCl LEFT OVER was titrated with NaOH.  So, you need to take the moles of NaOH used, which is equal to the moles HCl left over, and subtract that from 0.005 to find the moles of HCl THAT ACTUALLY REACTED.  Then use the stoichiometry to find moles of M.  I think it will be something like 0.00112 moles, or something like that.  Make sense?
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10/29/17

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