Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide.

H2(g) + CO2(g) --> H2O(g) + CO(g)

(b) Suppose 0.052 mol each of H2 and CO2 are placed in a 1.5 L container. When equilibrium is achieved at 986°C, what amounts of CO(g) and H2O(g), in moles, would be present? [Use the value of K calculated in part (a). (It was 1.6)]
[CO]
______ mol
[H2O]
______ mol

So I posted this before but was a little confused by the explanation. I got 2 answers, .066 and .044 but they are both wrong. Can someone explain?

To determine the amounts of CO(g) and H2O(g) present at equilibrium, we can use the given equilibrium constant (K) value and the initial amounts of H2(g) and CO2(g). Let's go through the calculations step by step:

Step 1: Calculate the initial moles of H2O(g) and CO(g):
Since the initial amounts of H2(g) and CO2(g) are given as 0.052 mol each, we can assume that the initial moles of H2O(g) and CO(g) are both zero.

[H2O]initial = 0 mol
[CO]initial = 0 mol

Step 2: Set up an ICE table:
An ICE table is a convenient way to keep track of the changes in moles of reactants and products at equilibrium. The table consists of three columns: Initial, Change, and Equilibrium.

For this reaction, the balanced equation is:
H2(g) + CO2(g) --> H2O(g) + CO(g)

ICE Table:
| H2(g) | CO2(g) | H2O(g) | CO(g)
-------------------------------------------------
Initial | 0.052 | 0.052 | 0 | 0
Change | -x | -x | x | x
Equilibrium| 0.052-x | 0.052-x | x | x

Step 3: Calculate the equilibrium concentrations of H2O(g) and CO(g):
We need to express the concentrations of H2O(g) and CO(g) in terms of moles per liter. The total volume of the container is given as 1.5 L.

[H2O] = x / 1.5
[CO] = x / 1.5

Step 4: Use the equilibrium constant expression to solve for x:
The equilibrium constant (K) for this reaction is given as 1.6. The expression for K is:

K = ([H2O] / [H2]) * ([CO] / [CO2])
1.6 = (x / 1.5) / (0.052 - x) * (x / 1.5) / (0.052 - x)

Simplifying the equation:
1.6 = (x^2) / ((0.052 - x)^2)

Step 5: Solve for x using the quadratic equation:
Rearranging the equation:
1.6 * (0.052 - x)^2 = x^2

Expanding and simplifying:
1.6 * (0.002704 - 0.104x + x^2) = x^2
0.0043264 - 0.1664x + 1.6x^2 = x^2

Rearranging again:
0 = 1.6x^2 - x^2 + 0.1664x - 0.0043264
0 = 0.6x^2 + 0.1664x - 0.0043264

Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(0.1664) ± √((0.1664)^2 - 4(0.6)(-0.0043264))) / (2(0.6))

Calculating x using the quadratic formula will give you the values for [H2O] and [CO] at equilibrium. Substituting the calculated value of x into the ICE table will give you the final amounts of CO(g) and H2O(g) in moles.

I apologize for the lengthiness of the calculation process, but it is necessary to accurately determine the equilibrium amounts of CO(g) and H2O(g) based on the given information and equilibrium constant.