Answer : The value of equilibrium constant (K) is, 0.004
Explanation :
First we have to calculate the concentration of 

and,

Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:

Initial conc. 1.2 0 0
At eqm. (1.2-2x) 2x x
As we are given:
Concentration of
at equilibrium = x = 0.1 M
The expression for equilibrium constant is:
![K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5E2%5BO_2%5D%7D%7B%5BSO_3%5D%5E2%7D)
Now put all the given values in this expression, we get:



Thus, the value of equilibrium constant (K) is, 0.004
<span>a. 0.325 g / 63.55 g/mol = 5.11 X 10^-3 moles Cu. SHould form 5.11 X 10^-3 mol Cu2+
b. Should form 5.11 X 10^-3 mol Cu(OH)2
c. 1 g Zn / 65.4 g/mol = 0.0153 mol Zn
Excess Zn = 0.0153 - 0.0051 = 0.0102 moles excess zinc
d. 5.11 X 10^-3 mol Mg X 24.3 g/mol = 0.124 grams Mg</span>
good sir it seems your answer appears to be B
Answer:
0.031moles
Explanation:
ok so it would help if I knew the marks but 1 mole = 1000L so if the volume was 31L then if you convert you will get that answer