Question

A solution initially contains 0.350 mol/L of A and 0.750 mol/L of B. A reaction occurs according to the equation
2 A(aq)+B(aq)<=> 3 C(aq)+D(s)
At equilibrium, [C] is found to be 0.100 mol/L. What is the value of Keq?

Answer 1

.........2A(aq) + B(aq) =>3C(aq) + D(s)
initial.0.350....0.750....0........0
change....-2x.....-x......+3x......x
equil...................0.100.......

If C is 0.100 M at equilibrium, then 3x = 0.100 and x = 0.100/3 = 0.0333
Then B must be 0.750-0.0333 = ? and A must be 0.350-2*0.0333 = ?
Keq = (B)/(A)^2*(B)
Substitute and solve for Keq. Note that D is not included because it is a solid.

Answer 2

thank-you so much for the help