I am  working on a chemistry problem and I'm stuck and need help. 
Your expertise is greatly appreciated.  Here is the problem:

1. Hydrochloric acid can dissolve solid iron according to the following reaction.

Fe(s) + 2 HCL (aq) -----> FeCL2(aq) + H2(g)

How much HCL in grams would you need to dissolve a 2.8-g iron bar on a
padlock? How H2 would be produced by the complete reaction on the iron
bar?

Thanks for your help!

Hi!!

From the equation you get that you need 2 moles of HCl to dissolve one
mol of Fe. And as a result you will get one mol of FeCl2 and one mol
of H2(g).

According to the Periodic Table of Elements one mol of Fe weighs
55.85g, for the purposes of this problem you can round it to 56g.
You have 2.8g of Fe, that is: 2.8/56 = 0.05 moles of Fe.

Now you know that for each mol of Fe you need two of HCl, then for
0.05 moles of Fe you will need 0.05*2 = 0.1 moles of HCl.

One mol of HCl weighs 36.5g aprox.; then 0.1 moles will weigh 3.65g.
This is the amount of HCl in grams that you need dissolve a 2.8g iron
bar.


Recall that for each mol of Fe dissolved you will get one mol of H2,
since 0.05 moles of Fe are dissolved, then 0.05 moles of H2 are
produced as a result of the reaction.
Since one mol of H2 weighs 2g, 0.05 moles of H2 will weigh 0.1g, and
this is the amount of H2 produced by the complete reaction on the iron
bar.


For references about molecular weight and mol weights see:
"Molecular mass - Wikipedia, the free encyclopedia":
http://en.wikipedia.org/wiki/Molecular_weight

"The Mole":
http://wine1.sb.fsu.edu/chm1045/notes/Stoich/Mole/Stoich04.htm

This page will be also useful for understanding this kind of problems
and many related:
"Reaction Equations" from the Faculty of Science Website of the
University of Waterloo, Ontario, Canada:
http://www.science.uwaterloo.ca/~cchieh/cact/c120/reaction.html

For a nice printable Periodic Table:
"Detailed Color Printable Periodic Table":
http://www.dayah.com/periodic/?lang=en



Search strategy:
molecular weight mol
periodic table
reactions moles weight


I hope this helps you. Feel free to request for a clarification if you need it.


Regards,
livioflores-ga