Question

N₂ (g) + 3 H₂ (g) 2 NH3 (9) If 28 g N₂ combines with 12 g of H₂ then maximum amount of NH3 form will be​

Answer 1

Answer:

∆Given equation: N2(g)+3H2(g)→2NH3

By observing the equation we see that 28gm of N2 and (3×2)=6gm H2 produced 2(14+3)=34gm of Ammonia/NH3. Here N2 is remain same but hydrogen is not equal so it is your answer↓

6gm of H2 produced 34gm of NH3

1gm ″ H2 ″ 34/6gm ″ NH3

12gm of H2 ″ =34/6×12″ NH3

=68gm

68 gm is the maximum amount of NH3.