Question

solve it fast pls i have to give exam ​

Answer 1

Answer:

1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g). ∴ 2000 g of N2 will react with H2 = 6/28 ×200g = 428.6g. Thus, here N2 is the limiting reagent while H2 is in excess. 28g of N2 produce 34g of NH3.