Question

when 14 gram h2 react with 6 gram n2 then calculate the volume of ammonia​

Answer 1

Answer:

The ba;lanced reaction is as follows:

N

2

+3H

2

→2NH

3

28 g 6 g 34 g

14 g 6 g (Given)

14 gm 3 gm (Actually consumed)

Here, nitrogen is limiting reagent so, excess hydrogen left is 3 g.

Answer 2

Answer:

N

2

+3H

2

⟶2NH

3

Number of moles of N

2

in 140g=

28

140g

=5 moles of N

2

Number of moles of H

2

in 30g=

2

30

=15 moles of H

2

moles N

2

+3H

2

⟶2NH

3

at t=0 5 15 -

after reaction 5−5×1 15−3×5 2×5

Thus 10 moles of NH

3

produced after reaction

as number of moles=

volume at STP

volume of product

⇒10=

22.4L

V

⇒V=224L