Question

28 gram of Nitrogen and 8 gram of hydrogen are mixed to produce Ammonia identify the limiting reagent ​

Answer 1

Answer:

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

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Explanation: