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Preparations and Uses of Sodium Bicarbonate and Common Salt

50 ml. of a m...

Question

50 ml. of a mixture of $N H_{3}$ and $H_{2}$ was completely decomposed by sparking into nitrogen and hydrogen. 40 ml of oxygen was then added and the mixture was sparked again. After cooling to the room temperature the mixture was shaken with alkaline pyrogallol and a contraction of 6 ml. was observed. Calculate the % of $N H_{3}$ in the original mixture. (Assuming that nitrogen does not react with oxygen.)

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Solution

Volume of mixture of $N H_{3}$ and $H_{2} V = 50 m l$
Let volume of $N H_{3}$ be V ml in mixture so hydrogen gas the volume of so - v ml
$2 N H_{3} \Rightarrow 2 N_{1} + 3 H_{2}$
So after decomposition of $N H_{3}$ we get V ml nitrogen $V \frac{3}{2}$ ml of hydrogen
total volume of $H_{2} = 50 - v + \frac{3 v}{2} = 50 + \frac{v}{2} m l$
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
$2 H_{2} + O_{2} =\Rightarrow 2 H_{2} O$
$50 + v / 2 m l$ of hydrogen react with $25 + c / 4$ ml of oxygen
The amount of oxygen left = $40 - (25 + v / 4)$
$= 15 - \frac{V}{4}$
when the mixture is passed through th alkaline pyrogallal $(C_{6} H_{3} (O H)_{3})$ the entire content of oxygen is absorbed it
Hence
$15 - \frac{v}{4} = 6$
$v = 36 m l$
Ammoma initially = $36 m l$
Percentage = $\frac{36 \times 100}{50} = 72 %$

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