Question

what will be amount of NH3 from when 22.4L of H2 and 44.8L of N2 react.​

Answer 1

Answer:

N

2

+H

2

⟶2NH

3

According to mole concept, 1 volume of nitrogen will react with 3 volumes of hydrogen to give 2 volumes of ammonia.

22.4 L of nitrogen will react with 67.2 L of hydrogen to give 44.8 L of ammonia

Explanation:

hope it helps

Answer 2

Answer: $22.4L$ $H$₂ and $44.8L$ $N$₂ reacted to form $14.93L$ $NH$₃

Explanation:

Given that,

The volume of $H$₂ = $22.4L$

The volume of $N$₂ = $44.8L$

The equation for the formation of ammonia is given by

$N$₂+ $3$$H$₂ ⇒ $2$ $NH$₃

$22.4L$ $N$₂+$67.2L$ $H$₂ ⇒ $44.8L$ $NH$₃

$1 mol = 22.4L$

Here, $H$₂  is the limiting reagent. $3$ $mol$ of $H$₂ gives $2$ $mol$ $NH$₃ gas but here only $1$$mol$  $H$₂ is present.

$3$ $mol$ of $H$₂ ⇒ $2$ $mol$ $NH$₃

$1$ $mol$ of $H$₂ ⇒ $\frac{2}{3}$ $mol$ $NH$₃

$22.4L$ of $H$₂ ⇒ $\frac{2}{3}$×$22.4$$L$ = $14.93L$ $NH$₃

∴ $22.4L$ $H$₂ and $44.8L$ $N$₂ reacted to form $14.93L$ $NH$₃.