N2 (g) + 3H2 (g) -->2 NH3 (g) The equation above is the equation for the Haber process. In a certain reaction, you start with 3.0 moles of nitrogen and 5.0 moles of hydrogen, which molecule is the limiting reagent?

Please please help! Im really struggling!

2 Answers
Apr 23, 2018

See below

Explanation:

Compare molar ratios:
5.0cancel("mols H"_2)*(1 "mol N"_2)/ (3 cancel("mol H"_2))= 1.7 "mol N"_2" needed"

Since 1.7 "mol N"_2" are needed to react, and there are 3.0 "mol N"_2" of available, N_2 is in excess and H_2 is limiting.

Hope that helps.

"H"_2 is the limiting reagent.

Explanation:

"N"_2 + "3H"_2 -> "2NH"_3

In above reaction

"N"_2 : "H"_2 = 1 : 3

"1.0 mol N"_2 requires "3.0 mol H"_2

If we have "3.0 mol N"_2 then for complete reaction we must have at least 3 × 3.0 = 9.0\ "mol H"_2. But, we have only "5.0 mol" of "H"_2. Therefore, "H"_2 is limiting reagent.

In another way,
If we have "5.0 mol H"_2 then for complete reaction we must have at least 1/3 × 5.0\ "mol N"_2 ≈ "1.7 mol N"_2. We have "5.0 mol N"_2. So, "N"_2 is the excess reagent.