Hi, I have a question, and I didn't know where to post it so I just decided to e-mail you!
A 0.790g sample of unknown metal completely reacted with 25.00mL of 0.50 M HCl (the HCL is in excess). The remaining solution was titrated with 4.15mL of 0.1M NaOH.
How many mol of NaOH was used?
How many mol of HCL was in excess?
How many mol of HCL was reacted witht he metal?
IDentify the metal as Mg Ca or Zn.
Could you give me some guidelines to follow? thanks
Well, that was a fun one, and it isn’t so bad if you visualize the problem, maybe drawing a picture of the experiment (I usually write “M” for an unknown metal). The steps that you already gave me are the best way to tackle the problem, actually.
I do have a slight problem with how the question is written, unless I am doing something silly.... Well, first you have to know—and most gen chem students are taught this—that many metals react with acid, reducing (donating electrons to) the hydrogen to make H2 gas, which oxidizes the metal. So we could write:
“M” means unknown metal:
M + HCl -----> MCl(#) + H2
This is unbalanced, and we so far do not know how many chlorides will become incorporated into the salt product. I intentially left it unbalanced for now, because we don’t know (until you read on in the question) whether or not the metal and H+ react in a 1:1 ratio.
But if you continue to read, down to to bottom of the question, you see the unknown metal could be either Zn, Ca, or Mg. Ca and Mg always lose two electrons to get a +2 charge, and Zn usually does, too. So this affects the equation:
M + 2 HCl -----> MCl2 + H2
We have to imagine this reaction takes place, and all the HCl reacts, and there is some leftover, which is neutralized by OH- in the following way:
OH- + H+ -----> H20
(ignore the sodium and chloride ions, they are not sticky in water, and will not adhere to anything here. They will float around as spectator ions.)
So we see the acid and base react in a 1:1 manner, so the moles of base that react will be EQUAL to the moles of acid that is in excess. Your question rightly asks for how many moles of base react.
Whenever you see the “M”, you may want to re-write it as moles per 1 L or moles per 1000 mL. Otherwise you can't cancel anything. Then the word “moles” pops out at you, and you know you almost have what you want---you just need to cancel the volume units. Just make sure the volume units match so they cancel—what I did below was to convert 4.15 mL to Liters.
0.1 mol NaOH/L x 0.00415 L = 0.000415 mol NaOH
This is equal to the amount of excess acid, since the excess acid reacted completely with this in a 1:1 ratio.
So this is the answer to the second question, too.
Excess HCl = 0.000415 mol
So we next need to figure out how many total moles of HCl was used, and subtract this above amount to see just how much reacted with the unknown metal.
Total HCl (again re-write M as moles per 1 L or moles per 1000 mL, your preference) Note that I like to change the every volume unit to Liters, for example, I change mL of HCl to L, and cancel that way.
0.025 L HCl x 0.50 moles HCl/1L = 0.0125 total moles HCl used.
Subtract from this the amount that was excess (I'm ignoring sig figs)
0.0125 moles HCl - 0.000415 moles excess = 0.0121 moles reacted with the metal.
OK! Now if we look at the very top equation we see that the acid reacts with the metal in a 2:1 ratio. Remember from stoichiometery you use the ratio of coefficients of A and B in a balanced equation to convert from moles of A to moles of B:
0.0121 moles HCl x 1 mol M/2 moles HCl = 0.00605 moles M
One good way to identifiy something is to look up its molar mass (atomic weight, formula weight, molecular weight, whatever you want to call it—it is the mass in grams of a mole of the substance.) But wait, we have moles, we have grams, we can calculate its molar mass, which is grams per 1 mole!
If you have grams and moles, you can always divide grams by moles to get the molar mass!
0.79 g / 0.00605 moles = 130.7 g/mole
Oh, dear, this is not the molar mass of Zn, Ca, or Mg. But if you assume the person who wrote the question forgot that these react in a 2:1 ratio, and ingore the stoichiometry step, it ends up having a mass of
0.79/ 0.0121 = 65.3g/mole, which is that of zinc’s.
I hope that helps!