Consider the equation A(aq) + 2B(aq) 3C(aq) + 2D(aq). In one experiment, 45.0 mL of 0.050 M A is mixed with 25.0 mL 0.100 M B. At equilibrium the concentration of C is 0.0410 M. Calculate K.

A and B dilute each other when mixed so the final solutions as they are mixed, but before they react is

0.050M x 45/70 = about 0.032M for A. but you should confirm this an all others that follow and do them more accurately than I.
0.100 M x 25/70 = about 0.0357M for B.

....A(aq) + 2B(aq) ==> 3C(aq) + 2D(aq)
I...0.032...0.036........0........0
C.....-x.....-2x........3x.......2x
E..0.032-x..0.036-2x....3x........2x

The problem tells you that at equilibrium (C) = 0.041M. That is 3x in the above; therefore, x = 0.041/3. That means D is 2x etc etc.
Plug these values into K expression and solve for K.

To calculate the equilibrium constant (K) for the given reaction, you need to find the molar concentrations of each species at equilibrium using the given data. Using the stoichiometry of the equation, we know that 1 mole of A reacts with 2 moles of B to produce 3 moles of C and 2 moles of D.

Let's start by finding the moles of A and B:

Moles of A = volume of A x molarity of A = 45.0 mL x 0.050 M = 2.25 mmol
Moles of B = volume of B x molarity of B = 25.0 mL x 0.100 M = 2.50 mmol

Now, let's use the stoichiometry of the equation to determine the number of moles of C at equilibrium:

Moles of C = moles of A x (3 moles of C / 1 mole of A) = 2.25 mmol x (3/1) = 6.75 mmol

Next, calculate the concentration of C at equilibrium:

Concentration of C = moles of C / total volume of the solution at equilibrium
= 6.75 mmol / (45.0 mL + 25.0 mL) = 6.75 mmol / 70 mL = 0.0964 M

Finally, let's use the concentration of C at equilibrium to calculate the equilibrium constant (K) using the following equation:

K = [C]^3 / ([A] x [B]^2)

Plugging in the values we calculated:

K = (0.0964)^3 / (0.050 x (0.100)^2)
= 0.00089178 / 0.0005
= 1.7836

Therefore, the equilibrium constant (K) for the given reaction is approximately 1.7836.