pH=2.7
<h3>Further explanation</h3>
Acetic acid = weak acid
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Ka = acid ionization constant
M = molarity
Ka for Acetic acid(CH₃COOH) : 1.8 x 10⁻⁵
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.222}\\\\=0.001998=1.998\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.222%7D%5C%5C%5C%5C%3D0.001998%3D1.998%5Ctimes%2010%5E%7B-3%7D)

It would go B. A. E. D. C.
Hope I helped!
Answer:
2 mol NO2
Explanation:
3NO2(g)+H2O(l)→2HNO3(l)+NO(g)
from reaction 3 mol 1 mol
given 11 mol 3 mol
for 3 mol NO2 ----- 1 mol H2O
for x mol NO2 ----- 3 mol H2O
3:x = 1:3
x = 3 *3/1 = 9 mol NO2
So, for 3 mol H2O are needed only 9 mol NO2.
But we have 11 mol NO2. So, NO2 is in excess, and
11 mol NO2 - 9 mol NO2 = 2 mol NO2 will be left after reaction.