chemistry
posted by raf .
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) 7.3
b) 0.34
c) 0.040
d) 0.14
e) none of these
the answer for this is suppose to be C but i am getting .000195 can any one show me how to do it

A = (45.0 mL x 0.05M/70 mL total) = 0.03214M.
B = (25.0 mL x 0.1M/70 mL total) = 0.03571M.
C = 0
D = 0
..............A + 2B ==> 3C + 2D
initial..0.03214..0.0371..0....0
change see below.
equil....................0.0410M
change:
(C) was 0 initially, must have changed by +0.0410M.
(D) must be at equilibrium 0.0410 x (2/3) = 0.02733M so it changed by +0.02733M
A must have changed by (0.0410/3) = 0.01367M. Just add initial + change (a negative number) to arrive at the equilibrium value.
B must have changed by (0.0410/3)*2 = 0.02733M. Just add initial + change (a negative number) for equilibrium value.
Now plug in the numbers to the K expression. I obtained 0.03969 which rounds to 0.0397 to 3 significant figures and 0.04 looks to be the best choice. I expect we just rounded differently; you will notice i carried my numbers to four places.