The reaction of iron and water vapor results in an equilibrium reaction, 3 Fe(s) + 4 H2O(g) ¡ê Fe3O4(s) + 4 H2(g) and an equilibrium constant of 4.6 at 850¡ÆC. What is the concentration of Hydrogen present at equilibrium if the reaction is initiated with 28 g of H2O and excess iron, Fe, in a 10.0 L container?
mols H2O = grams/molar mass H2O = approx 1.6 but this is an estimate and you need a more accurate answer. I've estimated all of the others, too, so be sure to redo each step for better accuracy.
M H2O = mols/L = approx 0.16 M.
.....3Fes) + 4H2Og) ==> Fe3O4s) + 4H2g)
I.....xs......0.16M......0.........0
C.....xs......-4x........solid....4x
E.....xs....0.16-4x......solid....4x
K = 4.6 = (H2)^4/(H2O)^4
4.6 = (4x)^4/(0.16-4x)^4
Solve for x = (H2) at equilibrium.
Post your work if you get stuck.
0.184, but I don't know how I got it :( SOS please help
To find the concentration of hydrogen gas (H2) at equilibrium, we need to follow these steps:
Step 1: Convert the given mass of H2O to moles.
The molar mass of H2O is 18 g/mol.
Number of moles of H2O = 28 g / 18 g/mol = 1.56 mol
Step 2: Use the stoichiometry of the balanced equation to determine the ratio of H2 to H2O.
According to the balanced equation, 4 moles of H2O react to produce 4 moles of H2.
Step 3: Calculate the number of moles of H2 produced.
Number of moles of H2 = (1.56 mol H2O) x (4 mol H2 / 4 mol H2O) = 1.56 mol
Step 4: Calculate the concentration of H2 at equilibrium.
The volume of the container is given as 10.0 L.
Concentration of H2 = Number of moles of H2 / Volume of container
Concentration of H2 = 1.56 mol / 10.0 L = 0.156 M
Therefore, the concentration of hydrogen gas present at equilibrium is 0.156 M.
To find the concentration of Hydrogen (H2) at equilibrium, you need to use the given information about the reaction, equilibrium constant, and initial amount of water (H2O).
First, let's calculate the number of moles of water (H2O) present initially:
Number of moles = mass / molar mass
The molar mass of water (H2O) is 18 g/mol.
Number of moles of H2O = 28 g / 18 g/mol
Number of moles of H2O = 1.56 mol
According to the balanced equation, the stoichiometric coefficient of H2O is 4. This means that for every 4 moles of water, 4 moles of Hydrogen (H2) are produced.
Since the reaction is initiated with excess iron (Fe), all the water will be consumed in the reaction, resulting in the formation of 4 moles of Hydrogen (H2).
Now, calculate the concentration of Hydrogen (H2) at equilibrium:
Concentration of H2 = Number of moles of H2 / Volume of container
Number of moles of H2 = 4 moles (from the stoichiometry)
Volume of container = 10.0 L (given)
Concentration of H2 = 4 moles / 10.0 L
Concentration of H2 = 0.4 mol/L
Therefore, the concentration of Hydrogen (H2) at equilibrium is 0.4 mol/L.