Question

50g 0f N2 combines with 10g of H2 to form NH3 . 1. Identify the limiting reagent 2. find the mass of NH3 produced.

Answer 1

N2+3H2=2NH3
28gm. 6 gm. 34 gm
28 gm N2 require 6g.m
50g.m will require (6/28)×50=10.7g.m H2
but we have only 10gm of H2
so H2 is limiting reagent in rxn
H2 will. decide the formation of NH3
6gm of H2 produce = 34 gm
10 g.m. will produce=(34/6)×10=56.6gm NH3
--------------------------------
limiting reagent is H2
Amount of NH3 produce =56.6gm

Answer 2

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,
1 g of N2 react with 6/28 g of H2

50g of N2 react with 6×50/28g of H2

6×50/28 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

e.g mass of NH3 produced = 340/6 g
= 56.667 g