3A(g) + 2B(g)− >4C(g) Kc = 1.33 × 10^21. At this temperature, 1.60 mol of A and 3.60 mol of B are placed in a 1.00 L container, what are the concentrations of A, B, and C at equilibrium?

Question

I have found that B=2.54 and C=2.13. I thought that A was 0, but it is not. I need to find A.

Answer 1

May be easiest to use an ICE table:

3A(g) + 2B(g) ====> 4C(g)

1.6............3.6...................0............Initial

-3x.........-2x...................+4x...........Change

1.6-3x.....3.6-2x.............4x.............Equilibrium

Keq = 1.33x10²¹ = [C]⁴ / [A]³[B]²

1.33x10²¹ = (4x)⁴ /(1.6-3x)³(3.6-2x)²

I'd hate to have to do the math, but maybe you can. If you can solve for x, then [A] = 1.6 - 3x.