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Under standard conditions, three dry flasks are filled with pure ammonia, air and HCl in equal volume mixed gas, NO and O2 in volume ratio of 4 Under standard conditions, three dry flasks are filled with: dry pure NH3, HCl gas containing half air, mixed gas of NO2 and O2[ V (NO2)∶V (O2)]=4∶17. The fountain experiments were then performed separately, and the ratio of the mass concentration of the solution obtained in the three flasks was () A.2∶1∶2 B.1∶1∶1 C.5∶5∶4 D.
What's the problem?
The answer is C 5:5:4
C (mass concentration)= n (mass of solute)/V (solution volume)
Then NH3 HCl is very soluble in water ~ so their volume will be occupied by water
Let NH3(or HCL) volume be VL (can be reduced later)
Then 1MOL gas is 22.4L under standard conditions (Vm=22.4L/MOL)
So VL gas is (VL/22.4L) MOL
The volume of water is then VL when the flask is full of water
Then NH3+H20===NH3*H20 has 1MOLNH3 to generate 1MOL ammonia
Then (VL/22.4L) MOL ammonia
Then C (concentration of substance)= n/V =(V/22.4)/V =1/22.4 MOL/L ≈0.045 MOL/L
And N02 with 02
4N02+02+2H20===4HNO3
1 Volume 02 is consumed for each 4 volume N02 reduction but only 4 HN03s are generated
Then 02 in the question is much larger than N02, and the rest of N0 is not considered.
Let N02 volume be VL (can be reduced later)
Then 1MOL gas is 22.4L under standard conditions (Vm=22.4L/MOL)
So VL gas is (VL/22.4L) MOL
Water volume when entering water is 5/4VL
Then 4N02+02+2H20===4HNO3
Then (VL/22.4L) MOLHN03
Then C (concentration of substance)= n/V =(V/22.4)/5/4 V =1*4/22.4*5MOL/L
The resulting concentration ratio is 1/22.4:1/22.4:1*4/22.4*5
It's 1:1:4/5.
Streamline 5:5:4
Use a flask filled with ammonia gas for fountain test. When the water fills the entire flask, the concentration of ammonia in the flask is (calculated according to the standard condition)() A.0.045mol • L-1 B.1 mol•L-1 C.0.029mol • L-1 D. Not sure
If the volume of the flask is VL, the amount of ammonia is: VL
22.4L/mol≈0.045Vmol,
The volume of the solution after the reaction was VL, and the mass concentration of the ammonia water in the flask was:0.045Vmol
VL =0.045 mol/L,
Therefore, A.
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