If 1.80 mol of A, 4.20 mol of B, and 1.00 mol of C are initially placed in a 3.50 Liter container. Then, at equilibrium, it was observed that 4.50 mol of C was present. a. What is the equilibrium constant expression for this reaction.

b. Use an I-C-E table to determine the equilibrium concentrations for remaining species, A and B.

c. Calculate the equilibrium constant for the reaction.

a. Kc = [C]³ / [A] × [B]²

b. [A]e = 0.179 M; [B]e = 0.530 M

c. 42.7

Explanation:

There is a missing part:

For the following equilibrium reaction: A(aq) + 2B(aq) <---> 3C(aq)

a. What is the equilibrium constant expression for this reaction.

The equilibrium  constant (Kc) is:

Kc = [C]³ / [A] × [B]²

b. Use an I-C-E table to determine the equilibrium concentrations for remaining species, A and B.

First, we will determine the concentrations for which we have information.

[A]i = 1.80 mol/3.50 L = 0.514 M

[B]i = 4.20 mol/3.50 L = 1.20

[C]i = 1.00 mol/3.50 L = 0.286 M

[C]e = 4.50 mol/3.50 L = 1.29 M

The I-C-E table is:

         A(aq) + 2B(aq) <---> 3C(aq)

I         0.514      1.20           0.286

C          -x          -2x              +3x

E      0.514-x    1.20-2x      0.286+3x

We know that,

0.286+3x = 1.29

x = 0.335 M

Then,

[A]e = 0.514-x = 0.514-0.335 = 0.179 M

[B]e = 1.20-2x = 1.20-2(0.335) = 0.530 M

c. Calculate the equilibrium constant for the reaction.

We will use the values calculated in b.

Kc = [C]³ / [A] × [B]²

Kc = 1.29³ / 0.179 × 0.530² = 42.7

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

a compound is a molecule made of atoms from different elements.

a. by the combination of an unlimited number of elements