56 g of N2 reacts 9 g of H2 in a closed vessel to form NH3. how much N2 is left?
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Given: The reaction is N
2
+3H
2
⇄2NH
3
56g6g27.5kg
Initial no. of moles of H
2
=
2
6
=3mol
N
2
=
2.8
56
=2mol
Initial number of moles 2
N
2
+ 3
3H
2
⇄0
2NH
3
At equilibrium, (2−x)(3−3x) 2x
∴ Number of moles of NH
3
=
17
27.54
=1.62
i.e. 2x=1.62
x=0.81
∴ At equilibrium,
No. of moles of N
2
=2−x
=2−0.81
=1.19
No. of moles in H
2
=3(1−x)
=3(1−0.81)
=0.57
∴ Equilibrium concentration of ,
[N
2
]=
1
1.19
=1.19M
[H
2
]=
1
0.57
=0.57M
[NH
3
]=
1
1.62
=1.62M
∴ Equilibrium constant,
K
c
=
[N
2
][H
2
]
3
[NH
3
]
2
=
1.19×0.57
3
1.62
2
=11.9≈12mol
2
lit
2
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