Question

28 gram nitrogen react with 4 g hydrogen to form ammonia.identify limiting reagent and mass of ammonia and unreacted substance.​

Answer 1

Question:

If 28g of nitrogen is mixed with 12g of hydrogen to form ammonia as per the reaction N2 + 3H2-> 2NH3, which is the limiting reagent in this reaction?

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.

P.S.- In stoichiometric problems, always do calculations on the molar basis.