The system CO2(g) + H2(g)<->H2O(g) + CO(g) is at equilibrium at some temperature. At equilibrium a 4.00 L vessel contains 1.00 mole CO2, 1.00 mole H2, 2.40 moles H2O, and 2.40

moles CO. How many moles of CO2 must be
added to the system to bring the equilibrium CO concentration to 0.679 mol/L?
Answer in units of moles

........CO2 + H2 ==> H2O + CO

I......0.25..0.25....0.60..0.60
C........x....x......-x.....-x
E.....0.679+x etc.

Then solve for x = (CO2) in M, and M x L = mols.

The Ksp of Ni(CN)2 is 3.0 × 10−23. What is

the molar solubility of Ni(CN)2 in a 0.065 M KCN solution?
Answer in units of mol/L.

I found that the normal molar solubility is 1.957x10^-8, but I don't know how the KCN solution affects it.
Sig figs aren't accounted for.

To find out how many moles of CO2 must be added to the system to bring the equilibrium CO concentration to 0.679 mol/L, we need to use the equilibrium constant expression, Kc.

The equilibrium constant expression for the given reaction is:

Kc = ([H2O] * [CO]) / ([CO2] * [H2])

In the given chemical equation, the stoichiometric coefficient of CO2 is 1, which means the coefficient of CO is also 1.

We are given the equilibrium concentrations as:

[CO2] = 1.00 mol
[H2] = 1.00 mol
[H2O] = 2.40 mol
[CO] = 2.40 mol (at equilibrium)

We are also given the desired concentration of CO:

[CO] = 0.679 mol/L

Now, we can substitute these values into the equilibrium constant expression and solve for [CO2].

Kc = ([H2O] * [CO]) / ([CO2] * [H2])
0.679 = (2.40 * 2.40) / ([CO2] * 1.00)

Now, rearrange the equation to solve for [CO2]:

[CO2] = (2.40 * 2.40) / (0.679 * 1.00)
[CO2] = 5.76 / 0.679
[CO2] ≈ 8.49 mol

To bring the equilibrium CO concentration to 0.679 mol/L, approximately 8.49 moles of CO2 need to be added to the system.