When CO2(g) (0.009242 mol/L) and 0.009242 mol/L of H2(g) in a 460.0 L reaction vessel at 619.0 °C are allowed to come to equilibrium the mixture contains 1.327 mol of CO(g). What is the equilibrium concentration (mol/L) of H2(g)?

CO2(g)+H2(g) = CO(g)+H2O(g)

This one is the same as the one just before and both use the same principle as all of the others.

is this simply x = -1.327

therefore .009242-(-1.327) = 1.3362

then 1.3362 / 460.00 = .0029?

Note here that CO and H2 are given in mols/L so you need to convert to mols first. Then it's done the same way as the CS2 and CH4 problem before.

To find the equilibrium concentration of H2(g), we need to use the balanced chemical equation and the concept of equilibrium constant.

First, let's write the balanced chemical equation for the reaction:
CO2(g) + H2(g) = CO(g) + H2O(g)

The equilibrium constant expression for this reaction can be written as:
Kc = [CO(g) * [H2O(g)] / [CO2(g)] * [H2(g)]

Given that the initial concentration of CO2(g) is 0.009242 mol/L, the initial concentration of H2(g) is also 0.009242 mol/L, and the equilibrium concentration of CO(g) is 1.327 mol/L.

We can substitute these values into the equilibrium constant expression and solve for the unknown concentration of H2(g):

Kc = [CO(g)] * [H2O(g)] / [CO2(g)] * [H2(g)]
Kc = 1.327 mol/L * [H2O(g)] / 0.009242 mol/L * 0.009242 mol/L

Now, we can rearrange the equation to solve for the equilibrium concentration of H2(g):

[H2(g)] = (1.327 mol/L * [H2O(g)]) / (0.009242 mol/L * 0.009242 mol/L)

To find the equilibrium concentration of H2(g), we need to know the concentration of H2O(g), which is not given. Without the concentration of H2O(g), we cannot calculate the equilibrium concentration of H2(g).