Assume that you have 1.41mol of H2 and 3.50mol of N2. How many grams of ammonia (NH3) can you make, and how many grams of which reactant will be left over?

Question

3H2+N2→2NH3

Answer 1

A limiting reagent problem with an extra twist at the end.
Write and balance the equation
N2 + 3H2 ==> 2NH3

Using the coefficients in the balanced equation, convert mols N2 and mols H2 to mols NH3.
N2 first:
1.41 mol N2 x (2 mol NH3/1 mol N2) = 2.82 mols NH3 produced (if we had 1.41 mol N2 and all of the H2 we needed.)

H2 next:
3.50 mol H2 x (2 mol NH3/3 mol H2) = 2.33 mol NH3 produced (if we had 3.50 mol H2 and all of the N2 we needed.)

You obtained two values and you know that can't be right; one of them must be wrong. In limiting reagent problems the smaller values is ALWAYS the correct one and the reagent producing that value is the limiting reagent. Therefore, you will have 2.33 mol NH3 produced and H2 is the limiting reagent.
To convert mols NH3 to grams it is g = mols x molar mass..

To find the mols of the "other" reagent(the non-limiting reagent or the excess reagent), use the coefficients again.

3.5 mol H2 x (1 mol N2/3 mol H2) = 1.17 mol N2 used.
1.41-1.17 = 0.24 mols N2 left un-reacted.
That will be 0.24 x 28 = about 6.8 grams N2 remaining.

Answer 2

idnt noioo