Question

what is the limiting reagent when 15.0 g of nitrogen react with 32.1 g hydrogen?

Answer 1

first write out balanced equation.
N2(g) + 3H2(g)---->2NH3(g)

next figure out how many moles of each reactants .

(150.0 g N2)*(1 mol N2/28.0 g) = 5.357 mols N2

(32.1 g H2)*(1 mol H2/2.02 g H2) = 15.891 mols H2

Compare one to the other....

(5.357 mols N2)*(3 mols H2/1mol N2)= 16.071 mols H2

That means if we begin with 5.357 mols N2, we need 16.071 mols H2. We only have 15.891 mols H2, so H2 is the limiting reagent.

Answer 2

he didnt really note, but the limiting reagent is just what its name suggests it is the reagent that is used up first in a reaction.

It is important in chemistry for monitoring reactions (if the limiting reagent is undetectable we can assume the reaction has completed) and also planning out reactions and conditions.

Answer 3

So essentially what you do is,

1. Find the amount in moles of both reagents, using mole = grams/molecular mass
2. Figure out how many moles of product is formed using the amount of moles with reactant 1 and then with reactant 2. The reactant that gives you the least amount of moles of product is your limiting reagent (this makes sense because like I said it is used up first)
3. Run a TLC plate using the limiting reagent as your starting material :) <- (not an actual step that applies to you)

Answer 4

This involves gas as well so you couldnt do a TLC with it :|

Answer 5

If you have any questions feel free to ask

Answer 6

Steps to find the limiting reagent.
1st step: find the moles of both reactants- this is available.
2nd step: use the mole ratio to find moles of nitrogen from hydrogen and vice versa - this is the needed moles.
3rd step: which moles needed is > available? this will be your limiting reagent.
Remember the limiting reagent is always consumed in a reaction.