Question

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!

Answer 1

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.

Answer 2

`"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.