Answer:
a) The chemical reaction is given as:
![6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)](https://tex.z-dn.net/?f=6NaOH%28aq%29%2BAl_2%28SO_4%29_3%5Crightarrow%202Al%28OH%29_3%28s%29%2B3Na_2SO_4%28aq%29)
2.571 grams of aluminum hydroxide is precipitated.
Explanation:
a) The chemical reaction is given as:
![6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)](https://tex.z-dn.net/?f=6NaOH%28aq%29%2BAl_2%28SO_4%29_3%5Crightarrow%202Al%28OH%29_3%28s%29%2B3Na_2SO_4%28aq%29)
b)
Moles of NaOH = n
Volume of the NaOH = 185.5 mL = 0.1855 L( 1 mL =0.001 L)
Molarity of the solution = 0.533 M
![n=0.533 M\times 0.1855 mL=0.09887 mol](https://tex.z-dn.net/?f=n%3D0.533%20M%5Ctimes%200.1855%20mL%3D0.09887%20mol)
Moles of aluminum = n
15.8 g of aluminum sulfate per liter.
Volume of solution = 627 mL = 0.627 L (1 mL= 0.001 L)
Mass of aluminium sulfate in solution = 15.8 g/L × 0.627 L =9.9066 g
Moles of aluminum sulfate = ![\frac{9.9066 g}{342 g/mol}=0.02897 mol](https://tex.z-dn.net/?f=%5Cfrac%7B9.9066%20g%7D%7B342%20g%2Fmol%7D%3D0.02897%20mol)
Moles of NaOH = 0.09887 mol
According to reaction, 6 moles of NaOH reacts with 1 mole of aluminum sulfate, then 0.09887 moles of NaOH will recat with :
![\frac{1}{6}\times 0.09887 mol=0.01648 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%7D%5Ctimes%200.09887%20mol%3D0.01648%20mol)
This means that sodium hydroxide moles are in limiting amount.So, amount of aluminum hydroxide will depend upon moles of sodium hydroxide.
According to reaction, 6 moles of sodium hydroxide gives 2 moles of aluminium hydroxide, then 0.09887 moles of sodium hydroxide will give :
of aluminum hydroxide
Mass of 0.03296 moles of aluminum hydroxide:
0.03296 mol × 78 g/mol = 2.571 g
2.571 grams of aluminum hydroxide is precipitated.