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InterviewSolution
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50 " mL of " a mixture of `NH_3` and `H_2` was completely decomposed into `N_2` and `H_2` by sparking. 40 " mL of " `O_2` was then added and the mixture was sparked again. After cooling the mixture was shaken with alkaline pyrogallol and a contraction of 6 mL was observed. Calculate the percentage of `NH_3` in the original mixture. |
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Answer» `underset(x)(NH_3)+underset(y)(H_2)to(1)/(2)underset((x)/(2))(N_2)+(3)/(2)underset((3)/(2)x)(H_2)+underset(ymL)(H_2)` `x+y=50mL` `underset((3)/(2)x)((3)/(2)H_2)+underset((3)/(4)x)((3)/(4)O_2)to(3)/(2)H_2O` `underset(ymL)(H_2)+underset((y)/(2)mL)(1)/(2)O_2tounderset(ymL)(H_2O)`. [After shaki ng with pyrogallol solution, unreacted `O_2` was adsobed and 6 mL contraction was observed Initial volume of `O_2=40ml` unreacted `O_2=6mL`, Reacted `O_2=40-6=34mL`] From equation (ii) and (iii), `O_2` used `(3)/(4)x+(y)/(2)=34implies3x+2y=34xx4` Soving equation (i) and (iv) we get `{:(3x+2y=136),(x+y=50):}]impliesy=14,x=36` Percentage of `NH_3=(36)/(50)=100=72%` Total volume of organic compound `+O_2=95mL` Volume of compound `=20mL` Volume of `O_2` used `=95-20=75mL` |
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