Search for Answers to Your Chemistry Questions

If 0.72g of Al is reacted with 17.0mL of 2.4 M KOH, which one is the limiting reagent

A sample of 0.72g of AL is reacted with 17.0mL of 2.4 M KOH.  Which one is the limiting reagent? 
      I came up with the formula as such 2Al(s)+2KOH(aq)+6H2O(l) yields 2KAl(OH)4 (aq)+3H2(g). 
      I went ahead and decided that I needed to get the moles of Aluminum and KOH by such methods:  17.0mL=0.0170L KOH X (2.4mol KOH/ 1L KOH)=0.0408 mol KOH/2
      Then for aluminum: 0.72g Al/26.98g mol Al= 0.0267 mol Al/2 
My final ratio was 0.0204mol KOH/0.0134mol Al.  Since Al was the smallest number I decided that it was the limiting reagent.  Is that correct?  And how would one go about finding the theoretical yield of the product, KAl(OH)4 formed if say 1.0g of the Al reacted with the KOH?

i have a couple of questions here:

1) where did you get the reaction? although I have searched google and found this reaction to be true it is not one that would easily predicted. did you reference it somewhere?

2) why do you have /2 after mol KOH and mol Al

Well first off I knew that Al and K would combine.  However, there is nothing to combine with the OH.  So I used the rule for balancing acidic,basic solutions to this same deal and found that it part worked.  Using what I had left, I finally found a problem that indeed gave me the correct formula.  However for the /2 I don't really know...I thought that maybe since there are 2 moles of each in the formula that perhaps I should divide each by 2.  Though now that I think about it, I probably shouldn't.

Here's another thought for you, Aluminum will react with excess concentrated solution of KOH as follows

    2 Al (s) + 2 KOH (aq) + 6 H2O (l) ? 2 Al(OH)4 (aq) + 2 K+ (aq) + 3 H2 (g)

The Al(OH)4 complex and K+ are both soluble, but if the concentration of OH- is not in excess the Al(OH)3 will begin to precipitate out of solution and the K+ remains as a soluble ion.

This looks like the first step in the laboratory production of potassium aluminum sulfate dodecahydrate (alum). KAl(SO4)2-12H2O.

0.72g Al * 1 mol Al  = 0.0267 mol Al (1 extra sig fig)
              26.98g Al

0.0170 L * 2.4 mol HOH = 0.408 mol KOH (1 extra sig fig)
                  1 Liter

The ratio of Al to KOH is 1 to 1, therefore, KOH is in excess and Al is limiting. I hope I didn't step on anyone toes by jumping in here.

Otis

No not at all Otis, it is a forum for people to learn

On a simpler note using the equation you cited above, Desert Lord

this is a typical limiting reactant problem

the first thing you will want to do is convert both reactants to moles of each reactant respectively

then pick just one product and solve for moles of that

you will have to use a mole ratio which is found from the balanced chemical equation

so

#moles of Al  x 3 mole H2 / 2 mole Al  = # moles of H2 produced

#moles of KOH x 3 moles H2 / 2 mole KOH = # moles of H2 produced

which ever reactant produces the smaller amount of moles of product , will be the limiting reactant

Alright, I understand.  Thank you both for cleaning that up.  I tend to over complicate my attempts at getting a figure =P




If you find this answer useful please share it with other students.