Answer: 1.
2. 3 moles of
: 2 moles of 
3. 0.33 moles of
: 0.92 moles of 
4.
is the limiting reagent and
is the excess reagent.
5. Theoretical yield of
is 29.3 g
Explanation:
To calculate the moles :

The balanced chemical equation is:
According to stoichiometry :
3 moles of
require = 2 moles of
Thus 0.33 moles of
will require=
of
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 3 moles of
give = 2 moles of
Thus 0.33 moles of
give =
of
Theoretical yield of
Thus 29.3 g of aluminium chloride is formed.
A homogeneous mixture is a mixture that is mixed so well together that you cannot see the different components that make up the mixture as the ingredients will not separate out, even over time.

Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

The mixture would contain
if
undergoes no hydrolysis; the solution is of volume
after the mixing. The two species would thus be of concentration
and
, respectively.
Construct a RICE table for the hydrolysis of
under a basic aqueous environment (with a negligible hydronium concentration.)

The question supplied the <em>acid</em> dissociation constant
for acetic acid
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant
for its conjugate base,
. The following relationship relates the two quantities:

... where the water self-ionization constant
under standard conditions. Thus
. By the definition of
:
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)


![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)
Cl2(g) -------> Cl-(aq) + ClO-(aq)
2e- + Cl2(g) -------> 2Cl-(aq) [reduction]
4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)