A mixture of 0.10 mole NO, 0.050 mole H₂, and 0.10 mole H₂O is placed in a 2.0 L vessel. The following equilibrium is established: 2 NO + 2 H₂(g) = N2(g) + 2 H₂O(g) At equilibrium [NO] = 0.030 M. The initial concentration of N₂ was not mentioned so you must assume [N₂] = 0. Calculate the equilibrium concentrations (in M) of: More Information X H₂(g) XN₂ (g) X H₂O (g) Question Answer Your answer -.020 is Your answer .035 is in Your answer .170 is i Tol: 0.0001 Tol: 0.001 Tol: ±0.001

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A mixture of 0.10 mole NO, 0.050 mole H₂, and 0.10 mole H₂O is placed in a 2.0 L vessel. The following equilibrium is established:
+ 2 H₂ (g) = N2(g) + 2 H₂O (g)
2 NO
At equilibrium [NO] = 0.030 M.
The initial concentration of N₂ was not mentioned so you must assume [N₂] = 0.
Calculate the equilibrium concentrations (in M) of:
More Information
X H₂ (g)
XN₂ (g)
X H₂O (g)
Question
Answer
Your answer -.020 is
Your answer .035 is in
Your answer .170 is i
Tol: 0.0001
Tol: 0.001
Tol: 0.001
Transcribed Image Text:A mixture of 0.10 mole NO, 0.050 mole H₂, and 0.10 mole H₂O is placed in a 2.0 L vessel. The following equilibrium is established: + 2 H₂ (g) = N2(g) + 2 H₂O (g) 2 NO At equilibrium [NO] = 0.030 M. The initial concentration of N₂ was not mentioned so you must assume [N₂] = 0. Calculate the equilibrium concentrations (in M) of: More Information X H₂ (g) XN₂ (g) X H₂O (g) Question Answer Your answer -.020 is Your answer .035 is in Your answer .170 is i Tol: 0.0001 Tol: 0.001 Tol: 0.001
Expert Solution
Step 1

Given ->

Initial mole of NO = 0.10 mole 

Initial mole of H2 = 0.050 mole 

Initial mole of H2O = 0.10 mole 

Volume= 2.0 L 

At equilibrium [NO] = 0.030 M 

 

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