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tommyq tommyq
wrote...
12 years ago
A strip of aluminum in an aqueous solution of hydrochloric acid will dissolve to create an aqueous solution of aluminum chloride with the release hydrogen gas.

What is the limiting reagent if a strip of aluminum (26.66 g) is placed in the solution which contains 3.025 mols hydrochloric acid?

Any help with this would be greatly appreciated, as I am stuck!!
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judypn Author
wrote...
12 years ago
First write the balanced chemical rxn:
2 Al(s) + 6 HCl(aq) Rightwards Arrow 2 AlCl3(aq) + 3 H2(g)

Now we can use stoichiometry to find out how much of a reagent we need. You can pick any of the reagents. I will choose the Al(s). I will now use stoichiometry to find out how much moles of HCl(aq) I need for 26.66 g of Al(s) to react with.

26.66 g Al x (1 mol / 26.98 g) x (6 mol HCl / 2 mol Al) = 2.964418087 mol HCl
We see that we only need 2.964418087 moles of HCl(aq) to react with 26.66 g of Al(s) that we have. But in the problem we have 3.025 moles of HCl(aq). This tells us that we have hydrochloric acid in excess of the amount needed. Therefore, HCl(aq) is the excess reagent (ER).
wrote...
12 years ago
First write the reaction between  HCl and Aluminum chloride. Dont forget to balance it!

Then  change the grams to mols and whichever produces less moles or grams of the product is the limiting reagent.

I'm posting a link under sources that shows a very clear example of how to set up the problem.
wrote...
12 years ago
Ok, first you have to balance this equation.

2Al + 6HCl Rightwards Arrow 2AlCl3 + 3H2 ta da!

The reason you need to do this is because the coefficients in the balanced equation tell you how much Al is needed to react with how much HCl and vice-versa.

Calculate the number of mols of aluminum: 26.66g/26.98 = .988 mol Al.

Now, the coefficients in the balanced reaction tell you how much HCl you need to react with this:

.988mol Al x (6mol HCl/ 2mol Al) = 2.964 mol HCl

This means that it takes 2.964 mol HCl to use up your aluminum. As a result, some HCl is left over, but you can't make anymore product because the aluminum is all gone. That makes aluminum the limiting reagent.

Note that if we had less than 2.964 mol HCl (instead of 3.025) we wouldn't have enough HCl to fully react with the aluminum. In that case, HCl would have been used up (making it the limiting reagent) and aluminum would have been left over.
wrote...
12 years ago
Write your equation:

Al(s) + HCl(aq)  ----->  AlCl3(aq) + H2(g)      (unbalanced)
Al(s) + 3HCl(aq)  ----->  AlCl3(aq) + 1.5H2(g)    (balanced)

STEP 1 for all these types of problems:   GET INTO MOLS!

26.66g x (1mol/26.98g) = .988mol Al(s)

.988mol Al
3.025mol HCl

STEP 2: Compare mol to mol ratio of actually and theoretical (coefficients in the equation).

Theoretical:   3mol HCl per 1mol Al    3/1=3
Actual: 3.025mol HCl per .988mol Al   3.025/.988=3.06

STEP 3: Use your human skill called logic for a second.  If you need 3mol HCl and have more then it is in excess.  If you need 1mol of Al but only have .988mol then it is limiting.  You can only make as much AlCl3 as you have of your limiting reactant.

Think about it in terms of eggs.  If you have 12 whole eggs, how many can you break.  Only 12..not 13 because you only have 12.  Mols is just a term for an amount of molecules.  1 dozen is 12 eggs like 1 mol is 6.022x10^23 molecules.

Answer: limiting reactant is Al(s)
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