A mixture of 0.10 molmol of NONO, 0.050 molmol of H2H2, and 0.10 molmol of H2OH2O is placed in a 1.0-LL vessel at 300 KK. The following equilibrium is established:

2NO(g)+2H2(g)←→N2(g)+2H2O(g)2NO(g)+ 2H2(g)←→N2(g)+2H2O(g)
At equilibrium [NO]=0.062M[NO]=0.062M.

Part A:

Calculate the equilibrium concentration of H2

Part B:

Calculate the equilibrium concentration of N2N2.

Part C:

Calculate the equilibrium concentration of H2OH2O.

Part D:

Calculate KcKc.

A) [H2] = 0.012 M

B) [N2]  = 0.019M

C) [H2O] = 0.138  M

D) Kc = 653.6

Explanation:

Step 1: Data given

Number of moles NO = 0.10 moles

Number of moles H2 = 0.050 moles

Number of moles H2O = 0.10 moles

Volume = 1.0 L

Temperature = 300 K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g)+2H2(g)←→N2(g)+2H2O(g)

Step 3: The initial  concentrations

Concentrations = moles / volume

[NO] = 0.10 moles / 1.0 L = 0.10 M

[H2] = 0.050 moles / 1.0 L = 0.050 M

[H2O] = 0.10 moles / 1.0 L = 0.10 M

[N2] = 0 M

Step 4: The concentration at the equilibrium

For 2 moles NO we need 2 moles H2 to produce 1 mol N2 and 2 moles H2O

[NO] = 0.10 - 2X

[H2] = 0.050 - 2X M

[H2O] = 0.10 + 2X M

[N2] = X

[NO] = 0.10 - 2X M = 0.062

X = 0.019

[NO] = 0.062M

[H2] = 0.050 - 2*0.019 = 0.012 M

[H2O] = 0.10 + 2*0.019 = 0.138  M

[N2] = X = 0.019M

Step 5: Calculate Kc

Kc = [N2][H2O]² / [NO]²[H2]²

Kc = (0.019*0.138²)/(0.062²*0.012²]

Kc = 653.6

explanation:

detials on each model with digrams

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"