Phosphorus pentachloride decomposes according to the chemical equation

I don't know what to do after I set up the Kc equation. How do you isolate x from 1.80=(x^2)/(.062-x)

First, I would check that 0.206/3.30. You shouldn't drop that last number; you are allowed three significant figures in 0.206 and 3.30 and that means you can have three s.f. in the answer. I have 0.0624. Second, your chemistry is right; you've done the hard part. Now it's just algebra. Here is what you start with.

(x^2)
--------- = 1.80
0.0624-x

Multiply both sides by 0.0624-x like this.
(0.0624-x)(x^2)
---------------- = 1.80*(0.0624-x)
(0.0624-x)

You see (0.0624-x) term on the left cancels since it is in the numerator and denominator and you are left with
x^2 = 1.80(0.0624-x)
Note: You could have done the same thing by cross multiplying but I don't know that you've seen that way.
Now multiply the right side to remove the parentheses, move all the terms to the left and you are left with a quadratic which you can solve for x. Let me know if you get stuck. Thanks for showing your work.

I was able to figure it out! Thank you very much for your help(:

To calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium, we can use the equilibrium constant expression and the given information.

Let's assign variables to the concentrations:
Let [PCl5] represent the equilibrium concentration of PCl5(g),
Let [PCl3] represent the equilibrium concentration of PCl3(g),
And let [Cl2] represent the equilibrium concentration of Cl2(g).

According to the balanced chemical equation:
PCl5(g) -----> PCl3(g) + Cl2(g)

The stoichiometry is 1:1:1, meaning:
For every 1 mole of PCl5 that decomposes, 1 mole of PCl3 and 1 mole of Cl2 are formed.

Given that the initial amount of PCl5(g) is 0.206 mol and the reaction vessel has a volume of 3.30 L, we can determine the initial concentration of PCl5(g):

[PCl5] = (0.206 mol) / (3.30 L) = 0.0624 M

Let's assume the change in concentration of PCl5(g) at equilibrium is -x, as it is being consumed. The change in concentration of PCl3(g) and Cl2(g) will also be +x, as they are being formed.

At equilibrium, the concentrations will be:
[PCl5] = 0.0624 - x (M)
[PCl3] = x (M)
[Cl2] = x (M)

We can now write the expression for the equilibrium constant (Kc) in terms of concentrations:

Kc = ([PCl3] * [Cl2]) / [PCl5]

Substituting the known values:
1.80 = (x * x) / (0.0624 - x)

Simplifying the equation:
1.80 * (0.0624 - x) = x^2

Expanding:
0.11232 - 1.80x = x^2

Rearranging the equation to set it equal to zero:
x^2 + 1.80x - 0.11232 = 0

Now, we can solve this quadratic equation for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

a = 1, b = 1.80, c = -0.11232

Calculating the discriminant:
D = b^2 - 4ac
D = (1.80)^2 - 4(1)(-0.11232)

D = 3.24 + 0.44864
D = 3.68864

Since the discriminant is positive, we have two solutions for x.

Using the quadratic formula:
x = (-1.80 ± √(3.68864)) / 2

Calculating x:
x₁ = (-1.80 + √(3.68864)) / 2
x₁ = (-1.80 + 1.92125) / 2
x₁ = 0.12125 / 2
x₁ = 0.060625

x₂ = (-1.80 - √(3.68864)) / 2
x₂ = (-1.80 - 1.92125) / 2
x₂ = -3.72125 / 2
x₂ = -1.860625

Since negative concentrations are not physically meaningful, we can discard the negative value. So, x = 0.060625 M.

Now, we can calculate the equilibrium concentrations:
[PCl5] = 0.0624 - x
[PCl5] = 0.0624 - 0.060625
[PCl5] = 0.001775 M

[PCl3] = x
[PCl3] = 0.060625 M

[Cl2] = x
[Cl2] = 0.060625 M

Therefore, at equilibrium:
[PCl5] = 0.001775 M
[PCl3] = 0.060625 M
[Cl2] = 0.060625 M