The decomposition of ammonia is characterized by the following decomposition equation:
2NH₃<span> → N</span>₂ <span> + 3H</span>₂
The mole ratio of N₂ : H₂ is 1 : 3
If the number of moles of N₂ = 0.0351 mol
Then the number of moles of H₂ = 0.0351 mol × 3
= 0.1053 mol
The number of moles of hydrogen gas produced when 0.0351 mol of Nitrogen gas is produced after the decomposition of Ammonia is 0.105 mol (OPTION 3).
Answer:
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(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???
when the reaction equation is:
C(s) + H2O(g) ↔ H2(g) + CO(g)
∴ Kc = [H2] [CO] / [H2O]
and we have Kc = 0.0393 (given missing in the question)
when the O2 is added so, the reaction will be:
2H2(g) + O2(g) → 2H2O(g)
that means that 0.15 mol H2 gives 0.15 mol of H2O
∴ by using ICE table:
[H2O] [H2] [CO]
initial 0.57 + 0.15 0 0.15
change -X +X +X
Equ (0.72-X) X (0.15+X)
by substitution:
0.0393 = X (0.15+X) / (0.72-X) by solving for X
∴ X = 0.098
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
= 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
= 0.72 - 0.098
= 0.622 M