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.
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.
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