50 ml of mixture of N*H_{3} and H_{2} are decomposed to yield N_{2} and H_{2} and to the resulting mixture 40 ml O_{2} is added and the mixture is sparked to yield H_{2}*O When the resulting mixture of gases were passed through alkaline pyrogallol 6 contraction was observed. Calculate the percentage composition of N*H_{3} in the original mixture.
Answers
Answer:
72 % of NH3
Explanation:
Volume of mixture of NH
3
and H
2
V=50ml
Let volume of NH
3
be V ml in mixture so hydrogen gas the volume of so - v ml
2NH
3
⇒2N
1
+3H
2
So after decomposition of NH
3
we get V ml nitrogen V
2
3
ml of hydrogen
total volume of H
2
=50−v+
2
3v
=50+
2
v
ml
this volume is less than 75 ml as v < 50 ml
Now 40 ml of O
2
is added. The amount of water formed and oxygen consumed is obtained as
2H
2
+O
2
=⇒2H
2
O
50+v/2ml of hydrogen react with 25+c/4 ml of oxygen
The amount of oxygen left = 40−(25+v/4)
=15−
4
V
when the mixture is passed through th alkaline pyrogallal (C
6
H
3
(OH)
3
) the entire content of oxygen is absorbed it
Hence
15−
4
v
=6
v=36ml
Ammoma initially = 36ml
Percentage =
50
36×100
=72%