Question

28g of N 2 is mixed with 12 g of H2 to form NH3.identify the limiting reagent​

Answer 1

Answer:

We can see that, for 1 mole of N2 we require 3 moles of H2 to produce 2 moles of NH3 and here for 1 mole of N2 we have 6 moles of H2 i.e. we have 3 moles of excess H2 which will be unused. Thus for the given case N2 will be the limiting reagent

Answer 2

Explanation:

Limiting reagent in a reaction is the one which is consumed totally before the total consumption of any other reactant and for the calculation of the same, it is better to do the calculations on mole basis.

So, for the given condition first let we calculate the number of moles of each reactant-

As we know,

Number of moles = Weight (in grams)/Molecular Wt.

Mol. Wt. of N2- 28

Mol. Wt. of H2- 2

Since, we have 28g of N2 & 12g of H2

Thus,

No. of moles of N2- 28/28 = 1 mole

& No. of moles of H2- 12/2 = 6 moles

Now, from the stoichiometry of the reaction

N2 + 3H2 → 2NH3

We can see that, for 1 mole of N2 we require 3 moles of H2 to produce 2 moles of NH3 and here for 1 mole of N2 we have 6 moles of H2 i.e. we have 3 moles of excess H2 which will be unused. Thus for the given case N2 will be the limiting reagent.

Thank you.