Question

a reaction mixture for the production of NH3 gas contains 250g of N2 gas and 50g of H2 gas under suitable conditions. Identify the limiting reactant, if any and calculate the mass of NH3 formed.

Answer 1

❤ hOlA mAtE ❤

==> N2 +3 H2 ---- > 2NH3

==> You can see that ,

==> 1 mole of N2 react with 3mole of H2 form 2mole of NH3 . e.g 28 g of N2 react with 6g of H2 form 34g of NH3.

==> SO ,

==> 1g of N2 react with 6/28 g of H2

==> 50g of N2 react with 6x50/28g of H2

==> 6x50/28 g > 10g of H2

==> Hence,it means H2 is limiting reagent because it is less then required amount .

==> Now ,

==> Reaction proceed based on limiting reagents .

==> Because , 6g of H2 form 34g NH3

==> So , 10g of H2 form 340/6 g of NH3

==> Example mass of NH3 produced = 340/6 g = 56.667 g

Hope it helps you ✔️