anonymous
  • anonymous
When 3.0 mol Al react with 7.0 mol HCl, what is the limiting reactant and how many moles of H2 can be formed? 2 Al + 6 HCl “yields”/ 2 AlCl3 + 3 H2 Al is the limiting reactant, 3.0 mol H2 can be formed HCl is the limiting reactant, 2.3 mol H2 can be formed Al is the limiting reactant, 4.0 mol H2 can be formed HCl is the limiting reactant, 3.5 mol H2 can be formed
Chemistry
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schrodinger
  • schrodinger
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anonymous
  • anonymous
The limiting reactant is the one that ends first and every single calculation is based on it. You can find out which one limits the total yield by comparing stoichiometric coefficients (for example the coefficient for HCl is this equation is 6). \[\frac{ n(HCl) }{ n(Al) } =\frac{ 6 }{ 3 }\] This is how you compare coefficients. For every six molecules/moles of HCl you'll need 3 molecules/moles of Al. If you play around with this simple equation (or proportion) you can find out how many moles of Al you'll need for every mole of HCl. Then compare the ratio to the amounts of substance you've been given, see which one ends first and calculate the rest using the limiting reactant.
anonymous
  • anonymous
So Al Is The Limiting Reactant?
anonymous
  • anonymous
Oh, sorry, I just noticed that I have a typo right there, it's supposed to be: \[\frac{ n(HCl) }{ n(Al) }=\frac{ 6 }{ 2 }\] (I am so sorry, don't know why I typoed that...) But if it had been like that then yes, Al would have been a limiting reactant. So I guess you grasped the theory itself even if I were a humongous idiot there? :D

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aaronq
  • aaronq
Instead of comparing the ratios, you can simply divide the amount of moles by it's coefficient, to get the "normalized" amount of moles (n). normalized moles=\(\dfrac{n_{species}}{coefficient}\) e.g. \(n_{normalized}=\dfrac{n_{Al}}{2}=\dfrac{3}{2}=1.5\;moles_{normalized}\) which ever has a lower value is obviously the limiting reactant.
anonymous
  • anonymous
xD Thank you!!

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