A mixture of 2.5 moles of H2O and 100 g of C are placed in a 50.0 L container and

allowed to come to equilibrium in the following reaction:
C(s) + H2O(g)  CO(g) + H2(g)
The equilibrium concentration of hydrogen gas is found to be 0.040 M. What is the
equilibrium concentration of the water vapour?

To find the equilibrium concentration of water vapor, we need to determine the moles of water present in the system first. We know that we have 2.5 moles of water in the mixture.

Next, we need to calculate the moles of hydrogen gas present in the system. Since we have the equilibrium concentration of hydrogen gas given as 0.040 M, we can use the ideal gas law to find the moles.

The ideal gas law equation is PV = nRT, where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of the gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
and T is the temperature in Kelvin.

Since we know the volume of the container is 50.0 L, we can rearrange the ideal gas law equation to solve for n:

n = PV / RT

We have the concentration (0.040 M) and volume (50.0 L) for hydrogen gas, so we can use the concentration formula C = n / V, where C is the concentration in moles per liter, to find the number of moles of hydrogen gas:

0.040 M = n / 50.0 L

n = 0.040 M * 50.0 L
n = 2.0 moles

Now that we have the moles of hydrogen gas, we can use the balanced equation of the reaction to determine the stoichiometric relationship between water vapor and hydrogen gas. From the balanced equation:

1 mole of C reacts with 1 mole of H2O to form 1 mole of H2.

Since we have 2.0 moles of H2, we can conclude that 2.0 moles of H2O have reacted.

Therefore, the equilibrium concentration of water vapor is:

Concentration = moles / volume
Concentration = (2.5 moles - 2.0 moles) / 50.0 L
Concentration = 0.1 M

Therefore, the equilibrium concentration of the water vapor is 0.1 M.