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.

50 ml of mixture of N*H_{3} and H_{2} are decomposed to yield N_{2} an

1.	50 ml of mixture of NH_{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.
Answer» 72 % of NH3Explanation: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%